A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the e

Question

3 years ago

6

A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the e

lectric potential due to the charge on the sphere is 18.0 V . Assume that V = 0 at an infinite distance from the sphere.
What is the electric potential at the center of the sphere?

Express your answer with the appropriate units.

Physics

2 answers:

gayaneshka [121] · Answer 1 · 2022-05-09T09:30:39+03:00

Answer:

V_inside = 36 V

Explanation:

Given

We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.

Required

We are asked to calculate the potential at the centre of the sphere

Solution

The potential energy due to the sphere is given by equation

V = (1/4*π*∈o) × (q/r) (1)

Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V

V ∝ 1/r

The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next

V_1/V_2=r_2/r_1

V_inside/V_outside = r/R

V_inside = (r/R)*V_outside (2)

Now we can plug our values for r, R and V_outside into equation (2) to get V_inside

V_inside = (1.2 m )/(0.600)*18

= 36 V

V_inside = 36 V