Question:

The electrostatic potential on the surface of uniformly charged spherical shell of radius R = 10 cm is 120 V. The potential at the centre of shell, at a distance r = 5 cm from centre, and at a distance r = 15 cm from the centre of the shell respectively, are:

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Remember that the electrostatic potential inside a uniformly charged spherical shell is constant and equal to the potential on the surface. Outside the shell, the potential varies inversely with the distance from the center.
Updated On: Oct 31, 2025
  • 120V, 120V, 80V
  • 40V, 40V, 80V
  • 0V, 0V, 80V
  • 0V, 120V, 40V
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The Correct Option is A

Approach Solution - 1

In this problem, we need to find the electrostatic potential at the center, at a distance of 5 cm from the center, and at a distance of 15 cm from the center of a uniformly charged spherical shell with a given radius \( R = 10 \, \text{cm} \). The surface potential is 120 V.

The concepts involved in this problem include:

  1. The electrostatic potential \( V \) at any point inside a uniformly charged spherical shell is constant and equal to the potential on the surface of the shell. This is a result of the shell theorem.
  2. Outside the shell, the potential \( V \) at a distance \( r \) from the center is given by the formula:

\(V = \frac{kQ}{r}\)

where \( k \) is Coulomb's constant and \( Q \) is the total charge on the shell.

Now, let's calculate the potential at each region mentioned:

  1. At the center of the shell:
    • According to the properties of a spherical shell, the potential at any point inside is the same as on the surface. Therefore, the potential at the center is 120 V.
  2. At a distance \( r = 5 \, \text{cm} \) from the center:
    • This point is inside the spherical shell (since \( 5 \, \text{cm} < 10 \, \text{cm} \)). Hence, the potential remains the same as on the surface. Thus, the potential is 120 V.
  3. At a distance \( r = 15 \, \text{cm} \) from the center:
    • This point is outside the spherical shell (since \( 15 \, \text{cm} > 10 \, \text{cm} \)). We use the formula for the potential outside the shell:

\(V = \frac{kQ}{15 \, \text{cm}}\)

where \( V = 120 \, \text{V} \, \text{at} \, r = 10 \, \text{cm} \). Therefore, the potential at 15 cm is less than the surface potential and is calculated by:

\(V_{\text{15 cm}} = \left(\frac{R}{15 \, \text{cm}}\right) \times 120 = \left(\frac{10}{15}\right) \times 120 = 80 \, \text{V}\)

Thus, the potentials are:

  • At the center: 120 V
  • At \( r = 5 \, \text{cm} \): 120 V
  • At \( r = 15 \, \text{cm} \): 80 V

The correct option is 120V, 120V, 80V.

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Approach Solution -2

To solve this question, we need to determine the electrostatic potential at three different positions relative to a uniformly charged spherical shell. These positions are:

  • At the center of the spherical shell.
  • At a distance of \( r = 5 \) cm from the center (inside the shell).
  • At a distance of \( r = 15 \) cm from the center (outside the shell).

The given information includes:

  • The radius of the sphere \( R = 10 \) cm.
  • The potential on the surface of the sphere is \( V = 120 \) V.

Now, let's analyze the potential at each point:

  1. Potential at the Center:
  2. Potential at \( r = 5 \) cm:
  3. Potential at \( r = 15 \) cm:

Therefore, the potentials at the distinct positions are:

  • At the center: 120 V
  • At \( r = 5 \) cm: 120 V
  • At \( r = 15 \) cm: 80 V

Hence, the correct answer is: 120V, 120V, 80V.

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