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Animated Solution for Physics - Electrostatics: A hollow metal sphere of radius $5 cm$ is charged such that the potential on its surface is $10 V$ . The potential at the centre of the sphere is

Q: I thought if the field is zero, the potential must be zero. Why isn't it?

Electric field is the *rate of change* of potential ($E = -dV/dr$). If a car's speed is zero, it doesn't mean its position is at the zero mile marker; it just means its position isn't changing. Similarly, $E=0$ means the potential isn't changing, so it remains at whatever value it had on the surface.

Select Answer:

Visualized Solution

\text{Visualizing the Charged Sphere}

\text{Radius, } R = 5\text{ cm}
V_{\text{surface}} = 10\text{ V}

\text{Electric Field Inside a Conductor}

E_{\text{inside}} = 0
E = -\frac{dV}{dr} \implies \frac{dV}{dr} = 0
V = \text{constant}

\text{Potential at the Centre}

V_{\text{centre}} = V_{\text{surface}} = 10\text{ V}

\text{What if it was a solid non-conducting sphere?}

V_{\text{centre}} = \frac{3}{2} V_{\text{surface}}

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The Sigma Insight: Electric Potential and Potential Difference

Solution Diagram

The Mystery of the Hollow Conductor

Imagine you are standing inside a hollow metallic sphere that has been charged. You might expect to feel a strong electric force, but surprisingly, you would feel absolutely nothing! This is one of the most beautiful and profound consequences of electrostatics: the electric field inside a hollow conductor is always zero, provided there are no charges enclosed within the cavity.

The Gradient of Potential

To understand why the potential at the center is

10 V

, we need to look at the relationship between the electric field (

E

) and the electric potential (

V

). The electric field is defined as the negative gradient of the potential:

E = - \frac{d V}{d r}

This equation tells us that the electric field measures how rapidly the potential changes as you move through space.

The Equipotential Volume

Since we know that

E = 0

everywhere inside the hollow metal sphere, it immediately follows that:

\frac{d V}{d r} = 0

In calculus, if the derivative of a function is zero over a certain region, the function itself must be a constant in that region. Therefore, the electric potential

V

does not change as you move from the surface to the center, or anywhere else inside the sphere. The entire volume of the conductor is an equipotential volume.

The Final Conclusion

Since the potential is constant everywhere inside, the potential at the center must be exactly equal to the potential on the surface.

V_{centre} = V_{surface} = 10 V

It is a simple yet powerful concept that frequently appears in competitive exams. Always remember: for a charged conductor, the potential inside is not zero; it is constant and equal to the surface potential!

Common Traps & FAQs

Why is the electric field inside a hollow conductor zero?

In a conductor, free electrons move until they reach electrostatic equilibrium. They distribute themselves on the outer surface in such a way that their individual electric fields perfectly cancel each other out everywhere inside the material and the cavity.

Does this rule apply to solid metal spheres too?

Yes! For any solid conductor, the excess charge also resides entirely on the outer surface. Therefore, the electric field inside a solid metal sphere is also zero, and its internal potential is constant and equal to the surface potential.

What if the hollow sphere is not perfectly spherical?

The rule holds true regardless of the shape. The electric field inside any closed hollow conductor (with no charges inside the cavity) is zero, and the entire volume is an equipotential region.

I thought if the field is zero, the potential must be zero. Why isn't it?

Electric field is the *rate of change* of potential (

E = - d V / d r

). If a car's speed is zero, it doesn't mean its position is at the zero mile marker; it just means its position isn't changing. Similarly,

E = 0

means the potential isn't changing, so it remains at whatever value it had on the surface.

What if we placed a point charge inside the hollow cavity?

If you place a charge inside the cavity, the electric field inside will no longer be zero. The field lines from the point charge will extend outward, and the potential will vary depending on the distance from that internal charge.

Similar Questions

JEE Main 2020

LEVELJEE Main

Concentric metallic hollow spheres of radii $R$ and $4 R$ hold charges $Q_{1}$ and $Q_{2}$ , respectively. Given that, surface charge densities of the concentric spheres are equal. The potential difference $V (R) - V (4 R)$ is

(A)

\frac{3 Q _{2}}{4 π ε _{0} R}

(B)

\frac{3 Q _{1}}{16 π ε _{0} R}

(C)

\frac{3 Q _{1}}{4 π ε _{0} R}

(D)

\frac{Q _{2}}{4 π ε _{0} R}

JEE Main 2015

LEVELJEE Advanced

A uniformly charged solid sphere of radius $R$ has potential $V_{0}$ (measured with respect to $\infty$ ) on its surface. For this sphere, the equipotential surfaces with potentials $\frac{3 V _{0}}{2}, \frac{5 V _{0}}{4}, \frac{3 V _{0}}{4}$ and $\frac{V _{0}}{4}$ have radius $R_{1}, R_{2}, R_{3}$ , and $R_{4}$ respectively. Then,

* Multiple Correct Options

(A)

R_{1} = 0

and

R_{2} > (R_{4} - R_{3})

(B)

R_{1} \neq = 0

and

(R_{2} - R_{1}) > (R_{4} - R_{3})

(C)

R_{1} = 0

and

R_{2} < (R_{4} - R_{3})

(D)

2 R < R_{4}

LEVELJEE Main

A thin spherical conducting shell of radius $R$ has a charge $q$ . Another charge $Q$ is placed at the centre of the shell. The electrostatic potential at a point $P$ at a distance $R /2$ from the centre of the shell is

(A)

\frac{2 Q}{4 π ε _{0} R}

(B)

\frac{2 Q}{4 π ε _{0} R} - \frac{2 q}{4 π ε _{0} R}

(C)

\frac{2 Q}{4 π ε _{0} R} + \frac{q}{4 π ε _{0} R}

(D)

\frac{2 ( q + Q )}{4 π ε _{0} R}

JEE Main 2019

LEVELJEE Main

A solid conducting sphere, having a charge $Q$ , is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V$ . If the shell is now given a charge of $- 4 Q$ , the new potential difference between the same two surfaces is

(A)

- 2 V

(B)

2 V

(C)

4 V

(D)

V

JEE Advanced 1981

LEVELJEE Main

A charge $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ ( $> r$ ) such that the surface densities are equal. Find the potential at the common centre.

LEVELJEE Main

A solid conducting sphere having a charge $Q$ is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V$ . If the shell is now given a change of $- 3 Q$ , the new potential difference between the same two surfaces is

(A)

V

(B)

2 V

(C)

4 V

(D)

- 2 V

JEE Advanced 2006

LEVELJEE Main

For spherical symmetrical charge distribution, variation of electric potential with distance from centre is given in diagram. Given that $V = \frac{q}{4 π ε _{0} R _{0}}$ for $r \leq R_{0}$ and $V = \frac{q}{4 π ε _{0} r}$ for $r \geq R_{0}$ Then, which option(s) are correct

* Multiple Correct Options

(A)

Total charge within

2 R_{0}

q

(B)

Total electrostatic energy for

r \leq R_{0}

is zero

(C)

r = R_{0}

electric field is discontinuous

(D)

There will be no charge anywhere except at

r = R_{0}

JEE Advanced 2019

LEVELJEE Advanced

A thin spherical insulating shell of radius $R$ carries a uniformly distributed charge such that the potential at its surface is $V_{0}$ . A hole with a small area $α 4 π R^{2}$ ( $α ≪ 1$ ) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct ?

(A)

The ratio of the potential at the center of the shell to that of the point at

\frac{1}{2} R

from center towards the hole will be

\frac{1 - α}{1 - 2 α}

(B)

The magnitude of electric field at the center of the shell is reduced by

\frac{α V _{0}}{2 R}

(C)

The magnitude of electric field at a point, located on a line passing through the hole and shell's center on a distance

2 R

from the center of the spherical shell will be reduced by

\frac{α V _{0}}{2 R}

(D)

The potential at the center of the shell is reduced by

2 α V_{0}

JEE Main 2020

LEVELJEE Main

A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R$ ( $R > r$ ). If the surface charge densities on the two shells are equal, the electric potential at the common centre is

(A)

\frac{1}{4 π ε _{0}} \frac{( R + r )}{2 ( R ^{2} + r ^{2} )} Q

(B)

\frac{1}{4 π ε _{0}} \frac{( 2 R + r )}{( R ^{2} + r ^{2} )} Q

(C)

\frac{1}{4 π ε _{0}} \frac{( R + 2 r ) Q}{2 ( R ^{2} + r ^{2} )}

(D)

\frac{1}{4 π ε _{0}} \frac{( R + r )}{( R ^{2} + r ^{2} )} Q

JEE Main 2019

LEVELJEE Advanced

A charge $Q$ is distributed over three concentric spherical shells of radii $a, b, c$ ( $a < b < c$ ) such that their surface charge densities are equal to one another.\nThe total potential at a point at distance $r$ from their common centre, where $r < a$ would be

(A)

\frac{Q ( a ^{2} + b ^{2} + c ^{2} )}{4 π ε _{0} ( a ^{3} + b ^{3} + c ^{3} )}

(B)

\frac{Q ( a + b + c )}{4 π ε _{0} ( a ^{2} + b ^{2} + c ^{2} )}

(C)

\frac{Q}{4 π ε _{0} ( a + b + c )}

(D)

\frac{Q ( ab + b c + c a )}{12 π ε _{0} ab c}

Select Answer:

Visualized Solution

\text{Visualizing the Charged Sphere}

\text{Electric Field Inside a Conductor}

\text{Potential at the Centre}

\text{What if it was a solid non-conducting sphere?}

The Sigma Insight: Electric Potential and Potential Difference

The Mystery of the Hollow Conductor

The Gradient of Potential

The Equipotential Volume

The Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A charge Q is distributed over two concentric hollow spheres of radii r and R (>r) such that the surface densities are equal. Find the potential at the common centre.

A charge $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ ( $> r$ ) such that the surface densities are equal. Find the potential at the common centre.