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
\(\frac{Q}{4 \pi E_o E_r}\)
\(\frac{Q}{4 \pi E_o E_r (R +r)}\)
\(\frac{Q}{4 \pi E_o E_r(R - r)}\)
\(\frac{Q}{4 \pi E_o E_rR}\)
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The correct answer is B
treating the charge as if it were concentrated at the center for points outside the sphere. Total distance of the test charge p from the center of the sphere is r + R
To determine the electric potential at point \( p \) located at a distance \( r \) from the surface of an isolated metal sphere of radius \( R \) and carrying a charge \( Q \), situated in a medium of relative permittivity \( \varepsilon_r \), we can use the principles of electrostatics.
First, we note that the electric potential \( V \) at a distance \( r' \) from the center of a charged sphere in a medium with permittivity \( \varepsilon \) is given by:
\[ V(r') = \frac{1}{4 \pi \varepsilon} \frac{Q}{r'} \]
Here, \( \varepsilon \) is the permittivity of the medium, which can be expressed as:
\[ \varepsilon = \varepsilon_r \varepsilon_0 \]
where \( \varepsilon_0 \) is the permittivity of free space and \( \varepsilon_r \) is the relative permittivity of the medium.
Given that the test charge is placed at a distance \( r \) from the surface of the sphere, the total distance from the center of the sphere (denoted as \( r' \)) is:
\[ r' = R + r \]
Therefore, the electric potential \( V \) at the point \( p \)
