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 electric 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.

  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

36 V

Explanation:

The solid conducting sphere is a positive charge

and has radius R₁ = 0.6m

at a point R₂ = 1.20 m, the electric potential V = 18.0 V

V, electric potential = K q/R where k  = 1/4 πε₀

V is inversely proportional to R

V₁ = electric potential at the center

V₂ = electric potential at 1.2 m

then

V₁ /V₂ = R₂ / R₁

V₁ = V₂ ( R₂ / R₁) = 18.0 V ( 1.2 / 0.6 ) = 36 V

answer would be d.) i hope this

explanation:

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