Question

A spherical conductor of radius 12 cm has a charge of 1.6 × 10^–7C distributed uniformly on its surface. What is the electric field ?

1. inside the sphere
2. just outside the sphere
3. at a point 18 cm from the centre of the sphere?

Answer 1

(a) Radius of the spherical conductor, r = 12 cm = 0.12 m Charge is uniformly distributed over the conductor, q = 1.6 × 10⁻⁷ C Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it. 

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(b) Electric field E just outside the conductor is given by the relation,

\(E = \frac{1}{4πεº} . \frac{q}{r^2}\)

Where, ε_o = Permittivity of free space and \(\frac{1}{4πε_o}\) = 9 × 10⁹ Nm² C^-2

Therefore, \(E = \frac{9×10^9×1.6×10-7}{( 0.12)^2} =10^5NC^{-1}\)

Therefore, the electric field just outside the sphere is 10⁵ NC^-1.

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(c) Electric field at a point 18 m from the centre of the sphere = E₁

Distance of the point from the centre, d = 18 cm = 0.18 m

\(E_1 = \frac{1}{4πε_o}.\frac{q}{d_2} = \frac{9×109×1.6×10-7}{(1.8×10-2)2}=4.4×10^4 NC^{-1}\)

Therefore, the electric field at a point 18 cm from the centre of the sphere is 4.4 × 10⁴ NC^-1 .

Answer 2

(a) Electric Field Inside the Sphere
For a conductor in electrostatic equilibrium, the electric field inside the conductor is zero because the charges reside on the surface, and the field inside cancels out.

\(E_{\text{inside}} = 0 \, \text{N/C}\)

(b) Electric Field Just Outside the Sphere
To find the electric field just outside the surface of the sphere, we use Gauss's law. Consider a Gaussian surface just outside the sphere with radius \(R = 12 \, \text{cm} = 0.12 \, \text{m}\). The charge enclosed by this surface is \(Q = 1.6 \times 10^{-7} \, \text{C}\).
Gauss's law states:
\(\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}\)

For a spherical surface, the electric field E is uniform and radial, so:
\(E \cdot 4 \pi R^2 = \frac{Q}{\epsilon_0}\)

Solving for E:
\(E = \frac{Q}{4 \pi \epsilon_0 R^2}\)

Plugging in the values:
\(E = \frac{1.6 \times 10^{-7}}{4 \pi (8.85 \times 10^{-12}) (0.12)^2}\)

Let's calculate this:
\(E = \frac{1.6 \times 10^{-7}}{4 \pi \times 8.85 \times 10^{-12} \times 0.0144}\)

\(E = \frac{1.6 \times 10^{-7}}{1.601 \times 10^{-12}}\)

\(E =1.00 \times 10^{5} \, \text{N/C}\)
So the electric field just outside the sphere is:
\(E_{\text{outside}} = 10^{5} \, \text{N/C}\)

(c) Electric Field at a Point 18 cm from the Centre of the Sphere
For a point outside the sphere at a distance \(r = 18 \, \text{cm} = 0.18 \, \text{m}\) from the center, the charge enclosed is still \(Q = 1.6 \times 10^{-7} \, \text{C}\)We can treat the sphere as if all its charge were concentrated at its center.
The electric field at distance r is given by:
\(E = \frac{Q}{4 \pi \epsilon_0 r^2}\)

Plugging in the values:
\(E = \frac{1.6 \times 10^{-7}}{4 \pi (8.85 \times 10^{-12}) (0.18)^2}\)

Let's calculate this:

\(E = \frac{1.6 \times 10^{-7}}{4 \pi \times 8.85 \times 10^{-12} \times 0.0324}\)

\(E = \frac{1.6 \times 10^{-7}}{3.601 \times 10^{-12}}\)

\(E = 4.44 \times 10^{4} \, \text{N/C}\)

So the electric field at a point 18 cm from the center of the sphere is: \(E_{18\text{cm}} = 4.44 \times 10^{4} \, \text{N/C}\)