Question

An isolated sphere has capacitance 50pF.


1) calculate its radius.


2) how much charge should be placed on it to raise its potential to 10^4 V.

Answer 1

Given,

Capacitance = 50 pF

                     = $50*10^{-12} F$

V = $10^{4} V$

1) Radius = $\frac{1}{4\pi\epsilon_0}.C$

               = $9*10^{9}*50*10^{-12}$

                 =  $45*10^{-2} m$

                = $45 m.$

2) q = CV

        = $50*10^{-12}*10^{4}$

         = $5*10^{-7} C$

         

Answer 2

Capacitance :

Answer :

1. Radius = 45m

2. Charge, q = 50 × 10^-8 C

Explanation :

Refer the attached picture.