Question

Question 2.4: A spherical conductor of radius 12 cm has a charge of 1.6 × 10 −7 C distributed uniformly on its surface. What is the electric field (a) Inside the sphere (b) Just outside the sphere (c) At a point 18 cm from the centre of the sphere?

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-87

Answer 1

(a) $\bf{\text{inside the sphere,}}$
We know , charge have nature to reside outer surface of conductor. It means, charge inside the surface equals zero.
according to Gaussian theorem,
Ф = q/ε₀ , here q is charged inclosed the Gaussian surface.
∵ q = 0
so, Ф = 0 and flux , Ф = E.A = 0
so, E = 0
Hence , inside the sphere, electric field equals zero.

(b) $\bf{\text{just outside the surface}}$
Take a Gaussian surface of radius r > R = 12cm
then, charged inclosed into the Gaussian surface is q = 1.6 × 10⁻⁷ C
so, Ф = q/ε₀
so, EA = q/ε₀
E = q/ε₀A , here A is the surface area of Gaussian spherical surface
e.g., A = 4πr²
So, E = q/4πε₀r² = 9 × 10⁹ × 1.6 × 10⁻⁷/(12 × 10⁻²)²
= 10⁵ N/C

(C) $\bf{\text{at a point 18cm from the centre of the sphere}}$
Similarly explanation of (B),
So, E = kq/r²
Here , k = 9 × 10⁹ Nm²/C² , q = 1.6 × 10⁻⁷C and r = 18cm
So, E = 9 × 10⁹ × 1.6 × 10⁻⁷/(18 × 10⁻²)²
= 4.44 × 10⁴ N/C

Answer 2

$\mathfrak{\huge{\purple{\underline{Question:-}}}}$

A spherical conductor of radius 12 cm has a charge of 1.6 x 10-7C distributed uniformly on its surface. What is the electric field:

(A) Inside the sphere.

(B) Just outside the sphere.

(C) At a point 18 cm from the center of sphere.

$\mathfrak{\huge{\purple{\underline{Given:-}}}}$

➤ Radius of spherical conductor = 12 cm

➤ Charge is distributed uniformly over the surface = 1.6 × 10 −7 C

$\mathfrak{\huge{\purple{\underline{To \: Find:-}}}}$

➤ Electric field inside the sphere.

➤ Electric field just outside the sphere.

➤ Electric field at a point 18 cm from the center of sphere.

$\mathfrak{\huge{\purple{\underline{Solution:-}}}}$

Solution I:

Given,

Radius of spherical conductor is 12 cm, ie. 0.12 m

Charge is distributed uniformly over the surface is 1.6 × 10 −7 C

Electric field inside a spherical conductor is zero.

Solution II:

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

$\sf E=\dfrac{1}{4 \pi \epsilon_0} .\dfrac{q}{r^{2}}$

Here,

$\sf Permittivity \: of \: free \: space \: and \: \dfrac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} \: Nm^{2} \: C^{-2}$

Therefore,

$\sf E=\dfrac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}}$

Therefore, just outside the sphere the electric field is $\sf \underline{\underline{4.4 \times 10^{4} \: Nc^{-1}}}$

Solution III:

From the center of sphere the electric field at a point $\sf 18 \: m= E_1$

From the center of sphere the distance of point d = 18 cm = 0.18 m.

$\sf E_1=\dfrac{1}{4 \pi \epsilon_0} .\dfrac{q}{d^{2}}$

$\sf =\dfrac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{((1.8 \times 10)^{-2})^{2}}$

$\sf = \underline{\underline{4.4 \times 10^{4} \: Nc^{-1}}}$

$\mathfrak{\huge{\purple{\underline{Points \: to \: Note:-}}}}$

➤ The electrostatic potential on the perpendicular bisector due to an electric dipole is zero.

➤ Electrostatic potential is a state dependent function as electrostatic forces are conservative forces.

➤ Electrostatic potential due to a point charge q at any point P lying at a distance r from it is given by $\sf V=\dfrac{1}{4 \pi \epsilon_0} .\dfrac{q}{r}$