JEE Main 2012

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Animated Solution for Physics - Electrostatics: Consider a thin spherical shell of radius $R$ with its centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field $∣ E (r) ∣$ and the electric potential $V (r)$ with the distance $r$ from the centre, is best represented by which graph?

Q: Why is the electric field zero inside the shell?

According to Gauss's Law, the electric flux through a closed surface is proportional to the enclosed charge. For any spherical surface inside the shell, the enclosed charge is zero, hence the electric field is zero.

Q: Why isn't the potential zero inside if the electric field is zero?

Electric field is the negative gradient of potential ($E = -dV/dr$). If $E = 0$, it means the potential is not changing with distance, so it must be a constant, not necessarily zero. It takes the value it has on the surface.

Q: Why does the electric field graph have a sudden jump at $r=R$?

The surface of the shell has a surface charge density. When you cross a surface charge, the electric field experiences a discontinuity proportional to the surface charge density, while the potential remains continuous.

Q: Does the shell behave like a point charge for all points?

No, it only behaves like a point charge for points outside the shell ($r \ge R$). Inside, the behavior is completely different.

Q: What would happen if the shell was a solid non-conducting sphere with uniform volume charge?

In that case, the electric field inside would increase linearly from the center ($E \propto r$), and the potential would be a downward-opening parabola. The graphs would look very different!

Select Answer:

Visualized Solution

$Visualizing the Spherical Shell$

Consider a thin spherical shell of radius $R$ with a uniform positive surface charge density $σ$ .

$Electric Field Inside (r < R)$

According to Gauss's Law, the charge enclosed by any spherical Gaussian surface inside the shell is zero.
$\oint E \cdot d A = \frac{q _{in}}{ϵ _{0}} = 0$
$\Rightarrow E = 0 for r < R$

$Electric Field Outside (r \geq R)$

For points outside the shell, it behaves as a point charge located at its center.
$E = \frac{1}{4 π ϵ _{0}} \frac{q}{r ^{2}} for r \geq R$
$\Rightarrow E \propto \frac{1}{r ^{2}}$

$Electric Potential Inside (r \leq R)$

Since $E = 0$ inside the shell, no work is done in moving a charge within this region.
$E = - \frac{d V}{d r} = 0 \Rightarrow V = constant$
$V = \frac{1}{4 π ϵ _{0}} \frac{q}{R} for r \leq R$

$Electric Potential Outside (r \geq R)$

Outside the shell, the potential decreases inversely with distance $r$ .
$V = \frac{1}{4 π ϵ _{0}} \frac{q}{r} for r \geq R$
$\Rightarrow V \propto \frac{1}{r}$

$Matching with the Graphs$

The electric field is zero inside and decays as $1/ r^{2}$ outside.
The electric potential is constant inside and decays as $1/ r$ outside.
This perfectly matches the graphs shown in option (d).

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The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Solution Diagram

Analyzing the Setup

Imagine a thin spherical shell of radius

R

, carrying a uniform positive surface charge. We need to figure out how the electric field and potential vary as we move away from the center. This is a classic application of Gauss's Law and the relationship between electric field and potential.

The Electric Field Inside and Outside

Let's start with the electric field inside the shell. According to Gauss's Law, if we draw a spherical Gaussian surface inside the shell (

r < R

), the enclosed charge is zero.

\oint E \cdot d A = \frac{q _{in}}{ϵ _{0}} = 0

Therefore, the electric field everywhere inside the shell is exactly zero. The graph of

∣ E (r) ∣

stays flat at zero for

r < R

Now, for points outside the shell (

r \geq R

), the shell behaves as if its entire charge is concentrated at the center. So, the electric field follows the inverse-square law, jumping to a maximum at the surface and then decreasing as

1/ r^{2}

E = \frac{1}{4 π ϵ _{0}} \frac{q}{r ^{2}}

The Electric Potential Inside and Outside

What about the electric potential? Since the electric field inside is zero, no work is done in moving a charge within this region. The electric field is the negative gradient of the potential (

E = - d V / d r

). If

E = 0

, it means the potential doesn't change; it remains constant everywhere inside, equal to its value on the surface.

V = constant = \frac{1}{4 π ϵ _{0}} \frac{q}{R} for r \leq R

Outside the shell, the potential decreases inversely with distance

r

, just like a point charge.

V = \frac{1}{4 π ϵ _{0}} \frac{q}{r} for r \geq R

Conclusion

Comparing our derived graphs with the given options, we can clearly see that option (d) perfectly matches both the electric field and potential variations. The electric field is zero then decays, while the potential is constant then decays.

Common Traps & FAQs

Why is the electric field zero inside the shell?

According to Gauss's Law, the electric flux through a closed surface is proportional to the enclosed charge. For any spherical surface inside the shell, the enclosed charge is zero, hence the electric field is zero.

Why isn't the potential zero inside if the electric field is zero?

Electric field is the negative gradient of potential (

E = - d V / d r

). If

E = 0

, it means the potential is not changing with distance, so it must be a constant, not necessarily zero. It takes the value it has on the surface.

Why does the electric field graph have a sudden jump at

r = R

The surface of the shell has a surface charge density. When you cross a surface charge, the electric field experiences a discontinuity proportional to the surface charge density, while the potential remains continuous.

Does the shell behave like a point charge for all points?

No, it only behaves like a point charge for points outside the shell (

r \geq R

). Inside, the behavior is completely different.

What would happen if the shell was a solid non-conducting sphere with uniform volume charge?

In that case, the electric field inside would increase linearly from the center (

E \propto r

), and the potential would be a downward-opening parabola. The graphs would look very different!

Select Answer:

Visualized Solution

$Visualizing the Spherical Shell$

$Electric Field Inside (r < R)$

$Electric Field Outside (r \geq R)$

$Electric Potential Inside (r \leq R)$

$Electric Potential Outside (r \geq R)$

$Matching with the Graphs$

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Analyzing the Setup

The Electric Field Inside and Outside

The Electric Potential Inside and Outside

Conclusion

Similar Questions

A thin spherical shell of radius $R$ has charge $Q$ spread uniformly over its surface. Which of the following graphs most closely represents the electric field $E (r)$ produced by the shell in the range $0 \leq r < \infty$ , where $r$ is the distance from the centre of the shell?

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Spherical Shell

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field Inside (r<R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field Outside (r≥R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Potential Inside (r≤R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Potential Outside (r≥R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Matching with the Graphs

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Analyzing the Setup

The Electric Field Inside and Outside

The Electric Potential Inside and Outside

Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre

A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

$Visualizing the Spherical Shell$

$Electric Field Inside (r < R)$

$Electric Field Outside (r \geq R)$

$Electric Potential Inside (r \leq R)$

$Electric Potential Outside (r \geq R)$

$Matching with the Graphs$

A non-conducting solid sphere of radius $R$ is uniformly charged. The magnitude of the electric field due to the sphere at a distance $r$ from its centre