Advertisements
Advertisements
Question
Find the amount of work done in rotating an electric dipole of dipole moment 3.2 x 10- 8Cm from its position of stable equilibrium to the position of unstable equilibrium in a uniform electric field if intensity 104 N/C.
Advertisements
Solution
Given: p = 3.2 x 10- 8Cm, E = 104 N/C
To Find: Work done in rotating dipole (W)
Formula: W = pE (cos θ0 - cos θ)
Calculation:
At stable equilibrium, θ0 = 0°
At unstable equilibrium, θ = 180°
From formula,
W = pE (cos θ0 - cos θ)
= 3.2 × 10-8 × 104 (cos 0 - cos 180)
= 3.2 × 10-4 [1 - (-1)]
= 6.4 × 10-4 J
Work done in rotating an electric dipole is 6.4 × 10-4 J.
APPEARS IN
RELATED QUESTIONS
Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.
- Identify an equipotential surface of the system.
- What is the direction of the electric field at every point on this surface?
A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m2. Will he get an electric shock if he touches the metal sheet next morning?
What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
(Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Draw equipotential surfaces:
(1) in the case of a single point charge and
(2) in a constant electric field in Z-direction. Why are the equipotential surfaces about a single charge not equidistant?
(3) Can electric field exist tangential to an equipotential surface? Give reason
Why is there no work done in moving a charge from one point to another on an equipotential surface?
Draw the equipotential surfaces due to an electric dipole.
Write two important characteristics of equipotential surfaces.
S1 and S2 are the two imaginary surfaces enclosing the charges +q and -q as shown. The electric flux through S1 and S2 are respectively ______.

Assertion: Electric field is discontinuous across the surface of a spherical charged shell.
Reason: Electric potential is continuous across the surface of a spherical charged shell.
Consider the following statements and select the correct statement(s).
- Electric field lines are always perpendicular to equipotential surface.
- No two equipotential surfaces can intersect each other.
- Electric field lines are in the direction of tangent to an equipotential surface.
The diagrams below show regions of equipotentials.
(i)![]() |
(ii)![]() |
(iii)![]() |
(iv)![]() |
A positive charge is moved from A to B in each diagram.
Draw equipotential surfaces for (i) an electric dipole and (ii) two identical positive charges placed near each other.
Equipotential surfaces are shown in figure. Then the electric field strength will be ______.

What is meant by an equipotential surface?
A curve connecting points that all have equal electric potential is best described as:
The direction in which electric potential decreases is:
What is the shape of the equipotential surfaces for a single point charge (+q)?
For an electric dipole (+q, −q), the equipotential surfaces are best described as:
The relation between the electric field \[\vec{E}\] and the potential gradient is expressed as:
A particle floats between two parallel plates separated by 0.1 m with a potential difference of 4000 V. The particle carries a charge of \[1.6 \times 10^{-19}\] C. Taking \[g = 9.8\ \text{m/s}^2\], the mass of the particle is approximately:
A point charge of \[1\ \mu\text{C}\] produces a potential \[V = \frac{kq}{r}\]. At what distance from the charge does the potential equal 90 V? (Take \[k = 9 \times 10^9\ \text{N m}^2/\text{C}^2\])




