CBSE Class 12 · 2024 · ElectrostaticsLong Answer

(b)(i) A thin spherical shell of radius

R

has a uniform surface charge density

σ

. Using Gauss's law, deduce an expression for the electric field (i) outside and (ii) inside the shell. (ii) Two long straight thin wires AB and CD have linear charge densities

10 μ C/m

and

- 20 μ C/m

respectively. They are kept parallel to each other at a distance 1 m. Find the magnitude and direction of the net electric field at a point midway between them.

Solution

(i) Concentric Gaussian sphere of radius

r

.
Outside (

r > R

), enclosed charge

σ (4 π R^{2})

E = \frac{σ R ^{2}}{ε _{0} r ^{2}}

Inside (

r < R

), enclosed charge is zero:

E = 0

.
(ii) Midway, each wire is at

r = 0.5 m

.
Field of AB (away from AB):

E_{1} = \frac{2 k λ _{1}}{r} = \frac{2 \times 9 \times 1 0 ^{9} \times 10 \times 1 0 ^{- 6}}{0.5} = 3.6 \times 1 0^{5} V/m

Field of CD (towards CD):

E_{2} = \frac{2 \times 9 \times 1 0 ^{9} \times 20 \times 1 0 ^{- 6}}{0.5} = 7.2 \times 1 0^{5} V/m

Both point from AB towards CD, so they add:

E = 1.08 \times 1 0^{6} V/m, towards the negative wire CD

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CBSE Class 12 · 2024 · ElectrostaticsLong Answer

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