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
Answer:
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
