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प्रश्न
State Gauss’ Law.
Answer the following question briefly and to the point:
State Gauss’ theorem.
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उत्तर १
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.
उत्तर २
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`
संबंधित प्रश्न
A point charge causes an electric flux of −1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
- If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
- What is the value of the point charge?
Electric intensity outside a charged cylinder having the charge per unit length 'λ' at a distance from its axis is ________.
(a) E = `(2pi in_0 lambda)/(Kr^2)`
(b) E = `(in_0 lambda)/(2piKr^2)`
(c) E = `lambda/(2piin_0Kr)`
(d) E = `(4piin_0lambda)/(Kr^2)`
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.
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.
Use Gauss' law to derive the expression for the electric field `(vecE)` due to a straight uniformly charged infinite line of charge density λ C/m.
The electric field intensity outside the charged conducting sphere of radius ‘R’, placed in a medium of permittivity ∈ at a distance ‘r’ from the centre of the sphere in terms of surface charge density σ is
A spherical ball contracts in volume by 0.02% when subjected to a pressure of 100 atmosphere. Assuming one atmosphere = 105 Nm−2, the bulk modulus of the material of the ball is:
Through two parallel wires A and B, 10A and 2A of currents are passed respectively in opposite directions. If the wire A is infinitely long and the length of the wire B is 2m, then force on the conductor B, which is situated at 10 cm distance from A, will be:
Consider a sphere of radius R with charge density distributed as
ρ(r) = kr for r ≤ R
= 0 for r > R
- Find the electric field at all points r.
- Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
- Obtain the expression for the electric field intensity due to a uniformly charged spherical shell of radius R at a point distant r from the centre of the shell outside it.
- Draw a graph showing the variation of electric field intensity E with r, for r > R and r < R.
