- Gauss’s law is useful for finding the electric field in highly symmetric charge distributions (line, plane, sphere).
- For an infinitely long charged wire, the electric field is radial and depends only on the distance r from the wire.
- Electric field due to an infinite line charge decreases with distance:
E = \[\frac{\lambda}{2\pi\varepsilon_0r}\] - For an infinite plane sheet, the electric field is uniform and does not change with distance.
- Electric field due to an infinite plane sheet is:
E = \[\frac{\sigma}{2\varepsilon_0}\] - For a uniformly charged spherical shell, the field outside behaves like a point charge at the centre:
E = \[\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\] - Inside a uniformly charged spherical shell, the electric field is zero.
Definitions [25]
Definition: Electric Charge
The basic property of matter due to which it experiences electric force and shows attraction or repulsion, is called electric charge.
Definition: Electric Field
The space surrounding an electric charge q in which another charge q0 experiences a (electrostatic) force of attraction or repulsion, is called the electric field of the charge q.
OR
Electric field due to a charge Q at a point in space may be defined as the force that a unit positive charge would experience if placed at that point.
Define electric field.
The region in which the charge experiences an electric force is the electric field around the charge.
Definition: Electromagnetic Field
A time-dependent combination of electric and magnetic fields that propagates through space and can transport energy is called an electromagnetic field.
Definition: Electric Lines of Force
“An electric line of force is an imaginary smooth curve drawn in an electric field along which a free, isolated positive charge moves. The tangent drawn at any point on the electric line of force gives the direction of the force acting on a positive charge placed at that point.”
Define Electric Flux.
The number of electric field lines crossing a given area, kept normal to the electric field lines, is called electric flux.
Definition: Electric Dipole
An electric dipole is a pair of equal and opposite point-charges placed at a short distance apart.
Definition: Direction of Dipole Axis
“The line joining the two charges, pointing from the negative charge to the positive charge. This is known as the ‘direction of dipole axis’.”
Define electric dipole moment.
The electric dipole moment is defined as the product of the magnitude of one of the charges and the distance between the two equal and opposite charges.
Definition: Volume Charge Density
The charge per unit volume in a region of space, is called volume charge density.
\[\rho=\frac{\Delta Q}{\Delta V}\]
Definition: Continuous Charge Distribution
A charge distribution in which charge is treated as continuously spread over a line, surface, or volume (ignoring microscopic discreteness), is called continuous charge distribution.
Definition: Surface Charge Density
The charge per unit area on a surface, is called surface charge density.
\[\sigma=\frac{\Delta Q}{\Delta S}\]
Definition: Linear Charge Density
The charge per unit length along a line (such as a wire), is called linear charge density.
\[\lambda=\frac{\Delta Q}{\Delta l}\]
Definition: Radian
“1 radian is the angle which an arc of length equal to the radius of a circle subtends at the centre of the circle.”
Definition: Electric Flux
The electric flux is a measure of the number of lines of force passing through some surface held in the electric field. It is denoted by ФE.
OR
The electric flux through a surface is the dot product of the electric field and the area vector of the surface, is called electric flux.
Definition: Electric Flux Through a Surface
The electric flux linked with a surface in an electric field may be defined as the surface integral of the normal component of the electric field over that surface.
Definition: Electric Flux Density
In an electric field, the ratio of electric flux through a surface to the area A of the surface is called the 'electric flux density' at the location of the surface.
Mathematical Definition:
Electric flux density = \[\frac {Φ_E}{A}\]
For a plane surface normal to the electric field:
Electric flux density = \[\frac {E A}{A}\] = E
Definition: Steradian
“1 steradian is the solid angle subtended by a part of the surface of a sphere at the centre of the sphere, when the area of the part is equal to the square of the radius of the sphere.”
Definition: Plane Angle
The arc of a circle subtends an angle at the centre of the circle. This angle is called a 'plane angle'.
Definition: Conductors
Conductors are those through which electric charge can easily flow. Metals, human body, earth, mercury and electrolytes are conductors of electricity.
OR
Substances which offer high resistance to the passage of electricity and do not allow electricity to pass through them easily, are called insulators.
Definition: Insulators
Those substances in which electric charge cannot flow are called ‘insulators' (or dielectrics). Glass, hard-rubber, plastics and dry wood are insulators. Insulators have practically no free electrons.
OR
Substances which allow electricity to pass through them easily are called conductors.
Definition: Semiconductors
Substances whose resistance to the movement of charges is intermediate between conductors and insulators, are called semiconductors.
Definition: Elementary Charge
The smallest unit of electric charge, denoted by e, is called the elementary charge.
Definition: Point Charge
A charged body whose size is negligibly small compared to the distance between the charges under consideration, is called a point charge.
Define a unit charge.
One coulomb is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of 9.0 × 109 N.
Formulae [7]
Formula: Electric Field Due to a Point Charge
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}\]
Formula: Torque on a Dipole in a Uniform Electric Field
\[\vec τ\] = \[\vec p\] × \[\vec E\]
Magnitude: τ = pE sin θ
Formula: Electric Field Due to a Continuous Charge Distribution
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\sum\frac{\rho\Delta V}{r^{\prime2}}\hat{r}^{\prime}\]
Formula: Electric Field at a Point
E = \[\frac{1}{4\pi\varepsilon_{0}}\frac{q}{r^{2}}\] newton / coulomb
where \[\frac{1}{4\pi\varepsilon_{0}}\] = 9.0 × 109 newton meter2 / coulomb2.
Formula: Electric Flux Through a Flat Surface in a Uniform Field
ФЕ = E A cos θ
- If the plane surface is normal to the electric field (θ = 0):
ФЕ = ЕА cos 0 = ЕА - If the plane surface is parallel to the electric field (θ = 90°):
Or = E A cos 90° = 0 - For field lines entering the plane surface normally (θ = 180°):
ФЕ = ЕА cos 180° = -EA
Formula: Electric Flux
\[\Phi_{E}=\int_{A}\vec{\mathbf{E}}\cdot d\vec{\mathbf{A}}\]
where,
∫A = is the (surface) integral over the entire surface
ΦE = is positive when lines leave the surface, negative when they enter.
OR
Electric flux through a small area element:
dΦ = E ⋅ dS = E dS cos θ
Total electric flux through a surface:
Φ = \[\int_S\vec{E}\cdot d\vec{S}\]
Formula: Electric Field Due to an Electric Dipole
E = \[\frac{1}{4\pi\varepsilon_{0}}\frac{p}{r^{3}}\]
In vector notation:
\[\overrightarrow{\mathbf{E}}=-\frac{1}{4\pi\varepsilon_{0}}\frac{\overrightarrow{\mathbf{p}}}{r^{3}}\]
Theorems and Laws [5]
Law: Principle of Superposition of Electric Forces
Statement
The principle of superposition states that the net electric force acting on a given charge due to a number of other charges is equal to the vector sum of the individual forces exerted on it by each charge taken separately, assuming the other charges are absent.
Explanation / Mathematical Form
Consider a system of nnn point charges q1,q2,q3,…,qn.
The force acting on charge q1 due to the other charges is:
where
\[\vec F_{12}\] is the force on q1 due to q2,
\[\vec F_{13}\] is the force due to q3, and so on.
According to Coulomb’s law, the force on q1 due to q2 is:
\[\vec F_{12}\] = \[\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}^2}\hat{r}_{12}\]
Similarly, forces due to other charges can be written, and their vector sum gives the resultant force on q1.
Thus, the force between any two charges is independent of the presence of other charges.
Conclusion
The principle of superposition shows that:
- Electric forces obey vector addition.
- Each pair of charges interacts independently.
- The net force on a charge in a multi-charge system is found by adding all individual Coulomb forces vectorially.
State Gauss’s law on electrostatics and drive expression for the electric field due to a long straight thin uniformly charged wire (linear charge density λ) at a point lying at a distance r from the wire.
Gauss' Law states that the net electric flux through any closed surface is equal to `1/epsilon_0` times the net electric charge within that closed surface.
`oint vec" E".d vec" s" = (q_(enclosed))/epsilon_o`
In the diagram, we have taken a cylindrical gaussian surface of radius = r and length = l.
The net charge enclosed inside the gaussian surface `q_(enclosed) = lambdal`
By symmetry, we can say that the Electric field will be in radially outward direction.
According to gauss' law,
`oint vec"E".d vec"s" = q_(enclosed)/epsilon_o`
`int_1 vec"E" .d vec"s" + int_2 vec"E" .d vec"s" + int_3 vec"E". d vec"s" = (lambdal)/epsilon_o`
`int_1 vec"E". d vec"s" & int_3 vec"E". d vec"s" "are zero", "Since" vec"E" "is perpendicular to" d vec"s"`
`int_2 vec"E" . d vec"s" = (lambdal)/epsilon_o`
`"at" 2, vec"E" and d vec"s" "are in the same direction, we can write"`
`E.2pirl = (lambdal)/epsilon_o`
`E = lambda/(2piepsilon_o r)`
State Gauss’ Law.
The electric flux (ΦE) through any closed surface is equal to `1/in_0` times the ‘net’ change q enclosed by the surface.
ΦE = `oint vec E d vec A`
= `q/in_0`
∈0 = Permittivity of free space.
Gauss’ theorem states that the net electric flux over a closed surface is `1/epsilon_0` times the net electric charge enclosed by the surface.
Φ = `oint vec E * d vec A`
= `q/epsilon_0`
Theorem: Gauss' Theorem
Statement
Gauss’s theorem in electrostatics states that the total electric flux through any closed surface (called a Gaussian surface) is equal to \[\frac {1}{ε_0}\] times the net charge enclosed by the surface, irrespective of the shape and size of the surface.
Mathematical Form
ΦE = \[\oint\vec{E}\cdot d\vec{A}=\frac{q_{\mathrm{enc}}}{\varepsilon_0}\]
where
- \[\vec E\] = electric field intensity
- d\[\vec{A}\] = outward normal area element
- qenc = net charge enclosed
- ε0 = permittivity of free space
Proof (Outline)
Consider a point charge +q placed inside a closed surface.
The electric field at a point on the surface is
E = \[\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\]
The flux through a small area element dA is
dΦE = \[\vec{E}\cdot d\vec{A}=EdA\cos\theta\]
Since dAcosθ = r2dΩ,
dΦE = \[\frac{q}{4\pi\varepsilon_0}d\Omega\]
Integrating over the entire closed surface,
ΦE = \[\frac{q}{4\pi\varepsilon_0}\int d\Omega\]
But the total solid angle subtended by a closed surface is 4π,
ΦE = \[\frac {q}{ε_0}\]
Hence proved.
Key Note
Gauss’s law is most useful for symmetric charge distributions (spherical, cylindrical, planar) and is valid for all inverse-square law fields.
Law: Coulomb’s Law
Statement
Coulomb’s law states that the electrostatic force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges and is repulsive for like charges and attractive for unlike charges.
Explanation/Mathematical Form
Let two point charges q1 and q2 be placed at a distance r apart in vacuum (or air).
According to Coulomb’s law:
F ∝ q1q2
Combining the above relations:
F = k\[\frac {q_1q_1}{r^2}\]
where
F = electrostatic force between the charges,
r = distance between the charges,
k = proportionality constant.
In vacuum (or air),
k = 9.0 × 109 N m2C−2
Hence,
F = \[\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}\]
where ε0 is the permittivity of free space, given by
ε0 = 8.85 × 10−12 C2N−1m−2
If the charges are placed in a dielectric medium of permittivity ε,
F = \[\frac{1}{4\pi\varepsilon}\frac{q_1q_2}{r^2}\]
and since ε = Kε0,
F = \[\frac{1}{4\pi K\varepsilon_0}\frac{q_1q_2}{r^2}\]
where K is the dielectric constant of the medium.
Conclusion
Coulomb’s law quantitatively describes the force of attraction or repulsion between two point charges.
The force:
- depends on the magnitudes of charges,
- varies inversely as the square of the distance,
- acts along the line joining the charges, and
- decreases in a dielectric medium by a factor equal to its dielectric constant.
Key Points
Key Points: Concept of Charge
- Thales (≈2500 years ago) observed that amber rubbed with wool attracts light objects like paper and straw.
- William Gilbert (1600) showed that many materials, such as glass, ebonite, and sulphur, also show this effect.
- This attractive property is produced by rubbing (friction); a material showing it is said to be electrified, and the process is called frictional electricity.
- An electrified material possesses electric charge and is therefore called a charged body.
- Electric charge is quantised (q = ±ne, e = 1.6 × 10−19 C); there are two types of charges (positive and negative), as charges repel, unlike charges attract, and the SI unit of charge is coulomb (C).
Key Points: Electric Field
- A charge creates an electric field around it, even if no other charge is present.
- The electric field does not depend on the test charge used to measure it (if the test charge is very small).
- The field of a positive charge points outward; the field of a negative charge points inward.
- The strength of the electric field decreases as the distance from the charge increases.
- At equal distances from a point charge, the electric field has the same magnitude (spherical symmetry).
Key Points: Electric Field Due to a System of Charges
- The electric field due to many charges is the force on a unit test charge at that point.
- The total electric field is the vector sum of fields due to each charge (superposition principle).
- The electric field depends on the positions of the charges and changes from point to point in space.
Key Points: Physical Significance of Electric Field
- An electric field describes the electrical effect of a system of charges and does not depend on the test charge used to measure it.
- It is a vector quantity defined at every point in space and can vary from point to point.
- In changing situations, electromagnetic fields travel at the speed of light and can carry energy from one place to another.
Key Points: Properties of the Electric Lines of Force
- Electric field lines originate from positive charges and terminate on negative charges (or at infinity).
- The tangent to a field line at any point gives the direction of the electric field; in a uniform field, the lines are parallel and straight.
- No two electric field lines intersect, as this would imply more than one direction of the electric field at a point.
- Electric field lines do not pass through a conductor, showing that the electric field inside a conductor is zero.
- The density of field lines indicates field strength—closer lines represent a stronger field, while wider spacing represents a weaker field; the lines are continuous and imaginary, though the field is real.
Key Points: Applications of Gauss' Theorem
Key Points: Electric Flux
- SI unit of Electric flux = N·m²·C⁻¹ or V·m (since E = N / C = V/m).
- Dimensions of electric flux: [ΦE] = [E] [A] = [ML3T−3A−1]
Key Points: Applications of Gauss’ Theorem
- Point Charge:
Using a spherical Gaussian surface, the electric field due to a point charge is
E = \[\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\]which directly leads to Coulomb’s law. - Infinite Line of Charge:
For a uniformly charged infinite wire with linear charge density λ,
E = \[\frac{\lambda}{2\pi\varepsilon_0r}\]The field is radial and varies inversely with distance r. - Infinite Plane Sheet of Charge:
For a sheet with surface charge density σ,
E = \[\frac{\sigma}{2\varepsilon_0}\]The field is independent of distance from the sheet. - Two Parallel Charged Sheets:
The electric field is uniform between the sheets and zero outside when the sheets carry equal and opposite charges. - Charged Conductor:
The electric field inside a conductor is zero, and just outside the surface,
E = \[\frac{\sigma}{\varepsilon_0}\]where σ is surface charge density. - Uniformly Charged Spherical Shell / Conducting Sphere:
Outside the shell: behaves like a point charge at the centre
Inside the shell: the electric field is zero - Uniformly Charged Non-conducting Sphere:
Outside: E ∝ \[\frac {1}{r^2}\]Inside: E ∝ r, increasing linearly from centre to surface
Key Points: Gaussian Surface and its Properties
- A Gaussian surface is an imaginary closed surface used to calculate the electric flux of a vector field.
- It must be a closed surface (e.g., a sphere, cylinder, or cube); open surfaces such as discs or squares are not valid.
- The shape of the Gaussian surface should match the symmetry of the charge distribution so that the electric field is uniform or normal to the surface.
- The surface must not pass through any discrete charge, though it may pass through a continuous charge distribution.
- Electric flux through a Gaussian surface depends only on the charges enclosed, even though the electric field on the surface is due to both internal and external charges.
Key Points: Gauss' Theorem
- Gauss’ theorem establishes a connection between the electric flux through a closed surface and the charge enclosed, and is especially useful for highly symmetric charge distributions.
- An area can be treated as a vector quantity, with both magnitude (area) and direction (the outward normal to the surface).
- A solid angle is the three-dimensional analogue of a plane angle and describes how a surface appears from a point.
- The total solid angle subtended at a point by a closed surface, irrespective of its shape, is always the same.
- The solid angle subtended by an area element depends on its orientation and position relative to the point, being maximum when the area faces the point directly.
Key Points: Important Properties of Electric Charge
- Quantisation of charge: Electric charge exists in discrete packets, and the charge on any body is given by
q = ±ne
where n is an integer and e = 1.6 × 10−19 C is the elementary charge. - No fractional charge: Charge cannot exist as a fraction of eee (like 0.5e or 2.3e); hence, electric charge is atomic in nature.
- Conservation of charge: The total electric charge of an isolated system remains constant; charge can neither be created nor destroyed, only transferred.
- Experimental support: Processes such as rubbing, pair production and annihilation, and radioactive decay always conserve the net charge of the system.
- Invariance of charge: The value of electric charge does not change with velocity, unlike mass, which varies with speed.
Important Questions [100]
- Obtain an Expression for the Work Done to Dissociate the System of Three Charges
- A Hollow Cylindrical Box of Length 0.5 M and Area of Cross-section 25 Cm2 is Placed in a Three Dimensional Coordinate System as Shown in the Figure.
- A Hollow Cylindrical Box of Length 1 M and Area of Cross-section 25 Cm2 is Placed in a Three Dimensional Coordinate System as Shown in the Figure.
- Consider a System of N Charges Q1, Q2, ... Qn with Position Vectors → R 1 , → R 2 , → R 3 , ... ... → R N Relative to Some Origin 'O'.
- Show that If We Connect the Smaller and the Outer Sphere by a Wire, the Charge Q on the Former Will Always Flow to the Latter, Independent of How Large the Charge Q Is.
- Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. Predict the polarity of the capacitor.
- Why Do the Electrostatic Field Lines Not Form Closed Loops?
- Draw the Pattern of Electric Field Lines, When a Point Charge –Q is Kept Near an Uncharged Conducting Plate.
- The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of ______.
- Derive an expression for the electric field E due to a dipole of length '2a' at a point distant r from the centre of the dipole on the axial line
- Why Do the Electric Field Lines Never Cross Each Other?
- The Figure Shows the Field Lines on a Positive Charge. is the Work Done by the Field in Moving a Small Positive Charge from Q to P Positive Or Negative? Give Reason.
- Derive an Expression for the Electric Field Due to a Dipole of Dipole Moment → P at a Point on Its Perpendicular Bisector.
- Draw the Pattern of Electric Field Lines When a Point Charge +Q is Kept Near an Uncharged Conducting Plate.
- A Point Charge (+Q) is Kept in the Vicinity of an Uncharged Conducting Plate. Sketch the Electric Field Lines Between the Charge and the Plate?
- Two Charges of Magnitudes +4q and − Q Are Located at Points (A, 0) and (− 3a, 0) Respectively. What is the Electric Flux Due to These Charges Through a Sphere of Radius ‘2a’ with Its Centre
- The electric flux through a closed Gaussian surface depends upon ______.
- Define Electric Flux.
- Write S.I unit of electric flux.
- How Does the Electric Flux Due to a Point Charge Enclosed by a Spherical Gaussian Surface Get Affected When Its Radius is Increased?
- Find Out the Outward Flux to a Point Charge +Q Placed at the Centre of a Cube of Side ‘A’
- "The Outward Electric Flux Due to Charge +Q is Independent of the Shape and Size of the Surface Which Encloses Is."
- What is the Electric Flux Through a Cube of Side 1 Cm Which Encloses an Electric Dipole?
- Given the Electric Field in the Region E=2xi,Find the Net Electric Flux Through the Cube and the Charge Enclosed by It.
- Given a Uniform Electric Field E=5 x 10^3 i N/C , Find the Flux of this Field Through a Square of 10 cm on a Side Whose Plane is Parallel to the Y-z Plane. What Would Be the Flux Through the Same Square If the Plane Makes a 30° Angle with the X-axis ?
- Consider Two Hollow Concentric Spheres, S1 and S2, Enclosing Charges 2q and 4q Respectively as Shown in the Figure. (I) Find Out the Ratio of the Electric Flux Through Them
- Figure Shows Three Point Charges +2q, −Q and + 3q. Two Charges + 2q and −Q Are Enclosed Within a Surface ‘S’. What is the Electric Flux Due to this Configuration Through the Surface ‘S’?
- A Thin Straight Infinitely Long Conducting Wire Having Charge Density λ Is Enclosed by a Cylindrical Surface of Radius R And Length L, Its Axis Coinciding with the Length of the Wire.
- Given a Uniform Electric Filed → E = 4 × 10 3 ^ I N / C . Find the Flux of this Field Through a Square of 5 Cm on a Side Whose Plane is Parallel to the Y-z Plane.
- Two Charges of Magnitudes −3q and + 2q Are Located at Points (A, 0) and (4a, 0) Respectively. What is the Electric Flux Due to These Charges Through a Sphere of Radius ‘5a’ with Its Centre at
- Define Electric Flux.
- Two Charges of Magnitudes −2q and +Q Are Located at Points (A, 0) and (4a, 0) Respectively. What is the Electric Flux Due to These Charges Through a Sphere of Radius ‘3a’ with Its Centre a
- Given a Uniform Electric Field → E = 2 × 10 3 ^ I N/C, Find the Flux of this Field Through a Square of Side 20 Cm, Whose Plane is Parallel to the Y−Z Plane.
- Derive the Expression for the Electric Potential Due to an Electric Dipole at a Point on Its Axial Line.
- Obtain the Expression for the Torque τ Experienced by an Electric Dipole of Dipole Moment P in a Uniform Electric Field, E .
- In Which Orientation, a Dipole Placed in a Uniform Electric Field is in (I) Stable, (Ii) Unstable Equilibrium?
- An electric dipole of length 1 cm, which placed with its axis making an angle of 60° with uniform electric field, experience a torque of 6 √ 3 N m . Calculate the potential energy
- An Electric Dipole of Dipole Moment P Consists of Point Charges +Q and −Q Separated by a Distance 2a Apart. Deduce the Expression for the Electric Field E Due to the Dipole at a Distance X from the Centre of the Dipole on Its Axial Line in Terms of the Dipole Moment P
- Answer the Following Question. Derive an Expression for the Electric Field at Any Point on the Equatorial Line of an Electric Dipole.
- Drive the Expression for Electric Field at a Point on the Equatorial Line of an Electric Dipole
- Depict the Orientation of the Dipole in (I) Stable, (Ii) Unstable Equilibrium in a Uniform Electric Field.
- Depict the equipotential surfaces due to an electric dipole.
- An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 4 square root of 3 Nm. Calculate the potential energy of the dipole, if it has charge ±8 nC
- Define electric dipole moment.
- Define Dipole Moment of an Electric Dipole. is It a Scalar Or a Vector?
- Find the Resultant Electric Field Due to an Electric Dipole of Dipole Moment, 2aq, (2a Being the Separation Between the Charges ±± Q) at a Point Distant 'X' on Its Equator.
- An Electric Dipole of Length 2 Cm, When Placed with Its Axis Making an Angle of 60° with a Uniform Electric Field, Experiences a Torque of 8 √ 3 Nm. Calculate the Potential Energy of the Dipole
- Define Torque Acting on a Dipole of Dipole Moment → P Placed in a Uniform Electric Field → E Express It in the Vector from and Point Out the Direction Along Which It Acts.
- Define Electric Dipole Moment. is It a Scalar Or a Vector? Derive the Expression for the Electric Field of a Dipole at a Point on the Equatorial Plane of the Dipole.
- Write the Expression for the Torque → τ Acting on a Dipole of Dipole Moment → P Placed in an Electric Field → E .
- Deduce the Expression for the Torque → R Acting on a Planar Loop of Area → a Acting on a Planar Loop of Area → B . If the Loop is Free to Rotate, What Would Be Its Orientation in Stable
- Derive the expression for the torque acting on an electric dipole, when it is held in a uniform electric field. identify the orientation of the dipole in the electric field, in which it attains a
- An electric dipole of dipole moment 2 × 10-8 C-m in a uniform electric field experiences a maximum torque of 6 × 10-4 N-m. The magnitude of the electric field is ______.
- An electric dipole of length 2 cm is placed at an angle of 30° with an electric field 2 × 105 N/C. If the dipole experiences a torque of 8 × 10-3 Nm, the magnitude of either charge
- An electric dipole of dipole moment P→ is placed in a uniform electric field E→ with its axis inclined to the field. Write an expression for the torque T→ experienced by the dipole in vector form.
- Deduce the Expression for the Torque Acting on a Dipole of Dipole Moment in the Presence of a Uniform Electric Field
- A charge Q is placed at the centre of a cube. The electric flux through one of its faces is ______.
- State and Explain Gauss’S Law.
- A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in the Figure. What is the magnitude of the electric flux through the square? (Hint: Think of the
- A Charge ‘Q’ is Placed at the Centre of a Cube of Side L. What is the Electric Flux Passing Through Each Face of the Cube?
- A Thin Conducting Spherical Shell of Radius R Has Charge Q Spread Uniformly Over Its Surface. Using Gauss’S Law, Derive an Expression for an Electric Field at a Point Outside the Shell.
- Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.
- State Gauss’S Law for Magnetism. Explain Its Significance.
- Answer the Following Question. State Gauss'S Law for Magnetism. Explain Its Significance.
- State Gauss'S Law in Electrostatics. Show, with the Help of a Suitable Example Along with the Figure, that the Outward Flux Due to a Point Charge 'Q'. in Vacuum Within a Closed Surface
- State Gauss's law on electrostatics and drive expression for the electric field due to a long straight thin uniformly charged wire (linear charge density λ) at a point lying at a distance r from the
- An infinitely long positively charged straight wire has a linear charge density λ. An electron is revolving in a circle with a constant speed v such that the wire passes through the centre
- Draw a Graph to Show the Variation of E with Perpendicular Distance R from the Line of Charge.
- Use Gauss' law to derive the expression for the electric field (E→) due to a straight uniformly charged infinite line of charge density λ C/m.
- Explain Why, for a Charge Configuration, the Equipotential Surface Through a Point is Normal to the Electric Field at that Point
- If the Point Charge is Now Moved to a Distance 'D' from the Centre of the Square and the Side of the Square is Doubled, Explain How the Electric Flux Will Be Affected.
- A Point Charge Q is at a Distance of D/2 Directly Above the Centre of a Square of Side D, as Shown the Figure. Use Gauss' Law to Obtain the Expression for the Electric Flux Through the Square
- Draw a graph of kinetic energy as a function of linear charge density λ.
- Find the Work Done in Bringing a Charge Q from Perpendicular Distance R1 to R2 (R2 > R1)
- "For Any Charge Configuration, Equipotential Surface Through a Point is Normal to the Electric Field."
- An isolated point charge particle produces an electric field E→ at a point 3 m away from it. The distance of the point at which the field is E→4 will be ______.
- What is the Amount of Work Done in Moving a Point Charge Q Around a Circular Arc of Radius ‘R’ at the Centre of Which Another Point Charge ‘Q’ is Located?
- The Field Lines of a Negative Point Charge Are as Shown in the Figure. Does the Kinetic Energy of a Small Negative Charge Increase Or Decrease in Going from B to A?
- Why Must Electrostatic Field at the Surface of a Charged Conductor Be Normal to the Surface at Every Point? Give Reason?
- Consider two identical point charges located at points (0, 0) and (a, 0). Is there a point on the line joining them at which the electric field is zero?
- What is the Function of Uniform Radial Field and How is It Produced?
- A Charge is Distributed Uniformly Over a Ring of Radius 'a'. Obtain an Expression for the Electric Intensity E at a Point on the Axis of the Ring. Hence, Show that for Points at Large Distance from the Ring, It Behaves like a Point Charge.
- Consider two identical point charges located at points (0, 0) and (a, 0). Is there a point on the line joining them at which the electric potential is zero?
- A Point Object is Placed on the Principal Axis of a Convex Spherical Surface of Radius of Curvature R, Which Separates the Two Media of Refractive Indices N1 And N2 (N2 > N1).
- Using Gauss’S Law, Prove that the Electric Field at a Point Due to a Uniformly Charged Infinite Plane Sheet is Independent of the Distance from It.
- How is the Field Directed If (I) the Sheet is Positively Charged, (Ii) Negatively Charged?
- An Infinitely Large Thin Plane Sheet Has a Uniform Surface Charge Density +σ. Obtain the Expression for the Amount of Work Done in Bringing a Point Charge Q from Infinity to a Point, Distant R,
- Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances
- A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge`Q/2` is placed at its centre C and an other charge +2Q is placed outside the shell at a distance x from the centre
- Use Gauss'S Law to Find the Electric Field Due to a Uniformly Charged Infinite Plane Sheet. What is the Direction of Field for Positive and Negative Charge Densities
- Find the electric field intensity due to a uniformly charged spherical shell at a point, inside the shell. Plot the graph of electric field with distance from the centre of the shell.
- Find the Electric Field Intensity Due to a Uniformly Charged Spherical Shell
- A Small Conducting Sphere of Radius 'R' Carrying a Charge +Q Is Surrounded by a Large Concentric Conducting Shell of Radius Ron Which a Charge +Q Is Placed.
- Using Gauss’ Law Deduce the Expression for the Electric Field Due to a Uniformly Charged Spherical Conducting Shell of Radius R at a Point(I) Outside and (Ii) Inside the Shell.
- Using Gauss'S Law in Electrostatics, Deduce an Expression for Electric Field Intensity Due to a Uniformly Charged Infinite Plane Sheet.
- Plot a Graph Showing the Variation of Coulomb Force (F) Versus , ( 1 R 2 ) Where R is the Distance Between the Two Charges of Each Pair of Charges:
- Four Charges +Q, −Q, +Q And −Q Are to Be Arranged Respectively at the Four Corners of a Square Abcd of Side 'A'.
- Two Equal Balls with Equal Positive Charge 'Q' Coulombs Are Suspended by Two Insulating Strings of Equal Length. What Would Be the Effect on the Force When a Plastic Sheet is Inserted Between the Two?
- Three-point Charges Q, – 4q and 2q Are Placed at the Vertices of an Equilateral Triangle Abc of Side 'L' as Shown in the Figure. Obtain the Expression for the Magnitude of the Resultant Electric
- Write Any Two Important Points of Similarities and Differences Each Between Coulomb'S Law for the Electrostatic Field and Biot-savart'S Law of the Magnetic Field ?
Concepts [21]
- Electric Charge
- Electrical Conduction in Solids
- Principle of Superposition
- Electric Field
- Electric Field Due to a System of Charges
- Physical Significance of Electric Field
- Electric Lines of Force
- Electric Flux
- Electric Dipole
- Dipole in a Uniform External Field
- Continuous Charge Distribution
- Gauss’s Law
- Applications of Gauss' Theorem
- Charging by Induction
- Electric Field Intensity Due to a Point-Charge
- Uniformly Charged Infinite Plane Sheet and Uniformly Charged Thin Spherical Shell (Field Inside and Outside)
- Overview: Gauss' Theorem
- Conductors and Insulators
- Important Properties of Electric Charge
- Scalar Form of Coulomb’s Law
- Electric Field due to an Electric Dipole
