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NEET PHYSICSEasy

A hollow metal sphere of radius R is given +Q charges to its outer surface. The electric potential at a distance R/3 from the centre of the sphere will be:

A

\frac{1}{4\pi\varepsilon_0}\frac{Q}{9R}

B

\frac{3}{4\pi\varepsilon_0}\frac{Q}{R}

C

\frac{1}{4\pi\varepsilon_0}\frac{Q}{3R}

D

\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}

Step-by-Step Solution

  1. Properties of a Conductor: For a charged hollow metal sphere (a conductor), the charge resides entirely on the outer surface. The electric field inside the conductor is zero.
  2. Potential Inside: Since the electric field (E=−dV/drE = -dV/dr) is zero inside (r<Rr < R), the electric potential (VV) is constant throughout the interior volume and is equal to the potential at the surface.
  3. Calculation: The point in question is at a distance r=R/3r = R/3 from the center. Since R/3<RR/3 < R, the point lies inside the sphere. Therefore, Vinside=Vsurface=14πε0QRV_{inside} = V_{surface} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}. (This behavior is analogous to the gravitational potential inside a spherical shell discussed in Class 11 Physics, Chapter 8).
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