Question

A metal sphere of radius 15 cm has a net charge of 3.0 $ 10#8 C. (a) What is the electric field at the sphere’s surface? (b) If V ! 0 at infinity, what is the electric potential at the sphere’s surface? (c) At what distance from the sphere’s surface has the electric potential decreased by 500 V?

Answer 1

Answer:

E = 1.20 × $10^{4}$ N/C

V = 1800 V

x = 0.058 m

Explanation:

given data

radius = 15 cm

net charge = 3 × $10^{-8}$ C

electric potential decreased = 500 V

solution

we get here electric field at the sphere’s surface that is

electric field at the sphere’s surface E  =  $\frac{q}{4\pi \epsilon _o R^2}$   ............1

put here value

electric field at the sphere’s surface E  = $\frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15^2}$    

E = 1.20 × $10^{4}$ N/C

and

potential on surface of sphere is

V =  $\frac{q}{4\pi \epsilon _o R}$ ................2

V = $\frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15}$  

V = 1800 V

and

now we get distance that is x

and we know here

ΔV = V(x) - V   ..............3

substitute here value

-500V = $\frac{q}{4\pi \epsilon _o } \times (\frac{1}{R+x} - \frac{1}{R})$

-500 V = ${3\times 10^{-8}\times 8.99\times 10^9} \times (\frac{1}{0.15+x} - \frac{1}{0.15})$

solve it we get x

x = 0.058 m