Question

2.
5 सेमी त्रिज्या के चालक गोले पर 31-41 HC आवेश है। आवेश के पृष्ठ घनत्व की गणना करें।
1+1
A conducting sphere of radius 5 cm has a charge of 31.41 uc. Calculate the
surface density of charge.​

Answer 1

Given:

A conducting sphere of radius 5 cm has a charge of 31.41 uC.

To find:

Surface charge density on the sphere.

Calculation:

Surface charge density of the sphere is an index representing the distribution of charge over the surface area of the sphere.

Let surface charge density be denoted as $\sigma$.

$\boxed{ \sf{ \therefore \: \sigma = \dfrac{total \: charge}{total \: surface \: area} }}$

$\sf{ = > \: \sigma = \dfrac{31.41\times{10}^{-6}}{4\pi {r}^{2} } }$

$\sf{ = > \: \sigma = \dfrac{31.41\times{10}^{-6}}{4\pi {( \frac{5}{100} )}^{2} } }$

$\sf{ = > \: \sigma = \dfrac{31.41\times{10}^{-6}}{4\pi {( 5 \times {10}^{ - 2} )}^{2} } }$

$\sf{ = > \: \sigma = \dfrac{31.41\times{10}^{-6}}{4 \times \pi \times 25 \times {10}^{ - 4} } }$

$\sf{ = > \: \sigma = \dfrac{31.41\times{10}^{-6}}{4 \times 3.141\times 25 \times {10}^{ - 4} } }$

$\sf{ = > \: \sigma = \dfrac{10\times{10}^{-6}}{4 \times 25 \times {10}^{ - 4} } }$

$\sf{ = > \: \sigma = \dfrac{ {10}^{-1} }{4 \times 25 } }$

$\sf{ = > \: \sigma = {10}^{-3} \: C {m}^{ - 2} }$

So, final answer is :

$\boxed{\bold{ \: \sigma = {10}^{-3} \: C {m}^{ - 2} }}$