Question

A metal sphere of radius 11 cm has a net charge of 2.8 × 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 320 V?

Answer 1

Answer:

Therefore,

a) $E=20.82\ kN/C$

b) $V= 2290\ Volt$

c) distance from the sphere's surface = 1.8 cm

Explanation:

Given:

Radius, r = 11 cm = 0.11 m

Charge, $Q = 2.8\times 10^{-8}\ C$

To Find:

a) 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 320 V = ?

Solution:

Electric field at the sphere's surface is given as,

$E=\dfrac{k\times Q}{r^{2}}$

Where,

E = Electric Field,

$k = Coulomb's\ constant = 9\times 10^{9}$

Q = Charge

r  =  Radius

Substituting the values we get

$E=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{(0.11)^{2}}=20.82\ kN/C$

Now, Electric Potential at point surface is given as,

$V=\dfrac{k\times Q}{r}$

Substituting the values we get

$V=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{0.11}=2.29\ kV$

For distance from the sphere's surface has the electric potential decreased by 320 V,

So V becomes,

V = 2290 - 320 = 1970 Volt, then r =?

$V=\dfrac{k\times Q}{r}$

Substituting the values we get

$r=\dfrac{k\times Q}{V}\\\\r=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{1970}=0.128\ m$

Therefore the distance from the sphere's surface,

$d = 12.8 - 11 = 1.8\ cm$

Therefore,

a) $E=20.82\ kN/C$

b) $V= 2290\ Volt$

c) distance from the sphere's surface = 1.8 cm