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?

Therefore,

a)

b)

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

Explanation:

Given:

Radius, r = 11 cm = 0.11 m

Charge,

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,

Where,

E = Electric Field,

Q = Charge

r  =  Radius

Substituting the values we get

Now, Electric Potential at point surface is given as,

Substituting the values we get

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 =?

Substituting the values we get

Therefore the distance from the sphere's surface,

Therefore,

a)

b)

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

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)