An isolated metal sphere of radius R, carrying an electric charge Q, is situated in the medium of relative permitivity, Er. A test charge is placed at a point p, distance r from the surface of the sphere. Let Eo represent the permitivity of free space. The electric potential at p is given by the expression
A
\(\frac{Q}{4 \pi E_o E_r}\)
B
\(\frac{Q}{4 \pi E_o E_r (R +r)}\)
C
\(\frac{Q}{4 \pi E_o E_r(R - r)}\)
D
\(\frac{Q}{4 \pi E_o E_rR}\)
Verified Correct Answer: B.
Physical Principles & Numerical Solution
The electrical potential V at a point P located at a distance r from the surface of an isolated charged metal sphere is given by
V = \(\frac{\text{Q}}{4\pi\varepsilon}\).\(\frac{1}{\text{d}}\)
Where d = R + r is the distance from the center of the sphere to point P
\(\epsilon\) = \(\varepsilon_r\)\(\varepsilon_o\) is the permittivity of the medium, with \(\epsilon_r\) being the relative permittivity and \(\epsilon_o\) the permittivity of free space
\(\therefore\) V = \(\frac{\text{Q}}{4\pi \varepsilon_r \varepsilon_o}.\frac{1}{R + r}\)
Source: Myschool Past Questions Archive (Record #35205)