Question

9.
A spherical conductor of radius 10 cm has a charge
of 3.2 x 10-7 C distributed uniformly. What is
the magnitude of electric field at a point 15 cm
from the centre of the sphere?

(1) 1.28 x 105 N/C
(2) 1.28 x 106 N/C
(3) 1.28 x 107 N/C
(4) 1.28 x 104 N/C​

Answer 1

Answer:

according to me the answer is 2

Answer 2

Given:

A spherical conductor of radius 10 cm has a charge of 3.2 x 10-7 C distributed uniformly.

To find:

Magnitude of electric field intensity at a point 15 cm from the centre of the sphere.

Calculation:

Let the field intensity at the specified point be E;

When the specified distance r > Radius R:

$\therefore \: E = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{q}{ {r}^{2} } \bigg \}$

$= > \: E = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{3.2 \times {10}^{ - 7} }{ {( \frac{15}{100}) }^{2} } \bigg \}$

$= > \: E = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{3.2 \times {10}^{ - 7} }{ {( 0.15) }^{2} } \bigg \}$

$= > \: E = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{3.2 \times {10}^{ - 7} }{ 225 \times {10}^{ - 4} } \bigg \}$

$= > \: E = \dfrac{1}{4\pi \epsilon_{0} } \bigg \{ \dfrac{3.2 \times {10}^{ - 3} }{ 225 } \bigg \}$

$= > \: E = 9 \times {10}^{9} \times \bigg \{ \dfrac{3.2 \times {10}^{ - 3} }{ 225 } \bigg \}$

$= > \: E = 0.128 \times {10}^{6}$

$= > \: E = 1.28 \times {10}^{5} \:N{C}^{-1}$

So, final answer is:

$\boxed{ \bf{\: E = 1.28 \times {10}^{5} \:N{C}^{-1} }}$