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…

Given: A uniformly charged spherical shell has radius R = 10 cm and electrostatic potential on its surface VR = 120 V. Find: The potential at the centre, at r = 5 cm, and at r = 15 cm. For a uniformly charged spherical shell, the electrostatic potential at any point inside the shell is constant and equal to the potential on the surface. Therefore, V centre = 120 V and since 5 cm < 10 cm, the point at r = 5 cm is also inside the shell, so V5 cm = 120 V For a point outside the shell, the potential is the same as that due to a point charge at the centre: V = kQr At the surface, 120 = kQ10 cm At r = 15 cm, V15 cm = kQ15 cm = ( 1015) 120 = 80 V Inside a uniformly charged spherical shell, potential is the same everywhere and equals the surface potential. So the first two values are immediately 120 V and 120 V. Outside the shell, potential varies as 1r. Hence, V15 cm = 120 1015 = 80 V Therefore, the correct option is A.

The electrostatic potential on the surface of uniformly charged s… — Solution | JEE Physics

MCQEasyJEE 2025Electric Potential & Potential Energy

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

A
$120 \, \text{V}$ , $120 \, \text{V}$ , $80 \, \text{V}$
B
$40 \, \text{V}$ , $40 \, \text{V}$ , $80 \, \text{V}$
C
$0 \, \text{V}$ , $0 \, \text{V}$ , $80 \, \text{V}$
D
$0 \, \text{V}$ , $120 \, \text{V}$ , $40 \, \text{V}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A uniformly charged spherical shell has radius $R = 10 \, \text{cm}$ and electrostatic potential on its surface $V_R = 120 \, \text{V}$ .

Find: The potential at the centre, at $r = 5 \, \text{cm}$ , and at $r = 15 \, \text{cm}$ .

For a uniformly charged spherical shell, the electrostatic potential at any point inside the shell is constant and equal to the potential on the surface. Therefore,

V_{\text{centre}} = 120 \, \text{V}

and since $5 \, \text{cm} < 10 \, \text{cm}$ , the point at $r = 5 \, \text{cm}$ is also inside the shell, so

V_{5\text{ cm}} = 120 \, \text{V}

Outside Potential Calculation

For a point outside the shell, the potential is the same as that due to a point charge at the centre:

V = \frac{kQ}{r}

At the surface,

120 = \frac{kQ}{10 \, \text{cm}}

At $r = 15 \, \text{cm}$ ,

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

Use the Shell Property Directly

Inside a uniformly charged spherical shell, potential is the same everywhere and equals the surface potential. So the first two values are immediately $120 \, \text{V}$ and $120 \, \text{V}$ .

Outside the shell, potential varies as $\frac{1}{r}$ . Hence,

V_{15\text{ cm}} = 120 \times \frac{10}{15} = 80 \, \text{V}

Therefore, the correct option is A.

Common mistakes

Assuming that the potential inside the shell is zero because the electric field inside is zero is incorrect. Zero electric field means the potential is constant, not necessarily zero. Here it remains equal to the surface potential $120 \, \text{V}$ .
Using the outside formula $V = \frac{kQ}{r}$ for the point at $r = 5 \, \text{cm}$ is wrong because that point lies inside the shell. For all interior points of a charged spherical shell, use constant potential equal to the surface value.
Taking the potential at $r = 15 \, \text{cm}$ to vary as $\frac{1}{r^2}$ is incorrect. Potential outside a spherical shell varies as $\frac{1}{r}$ , while the electric field varies as $\frac{1}{r^2}$ .

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Answer

Step-by-step solution

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