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 \le r \lt \infty , where r is the distance from the center of the shell?

Electrostatics PYQ — JEE Main 2008 · Shift 0 · Q72 with Solution

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Correct answer: A

Use Gauss’s law for a uniformly charged thin spherical shell

For a spherical shell of radius $R$ carrying total charge $Q$ uniformly on its surface, symmetry tells us that the electric field depends only on the distance $r$ from the center.

Gauss’s law:

\oint \vec E \cdot d\vec A = \frac{Q_{\text{enclosed}}}{\varepsilon_0}

Region 1: Inside the shell ( $0 \le r < R$ )

Take a spherical Gaussian surface of radius $r<R$ .

Since all the charge lies on the shell at $r=R$ , the enclosed charge is

Q_{\text{enclosed}}=0.

So,

E(4\pi r^2)=0 \quad \Rightarrow \quad E=0.

Thus,

E(r)=0 \qquad (r<R).

Region 2: Outside the shell ( $r>R$ )

Now take a spherical Gaussian surface of radius $r>R$ . Then the full charge $Q$ is enclosed.

So,

E(4\pi r^2)=\frac{Q}{\varepsilon_0}

which gives

E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} \qquad (r>R).

So outside the shell, the field decreases as $1/r^2$ .

Behavior at $r=R$

Just inside the shell,

E(R^-)=0.

Just outside the shell,

E(R^+)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}.

So there is a sudden jump (discontinuity) at $r=R$ .

Shape of the graph

Therefore the correct graph must show:

$E=0$ for all $0\le r<R$
a jump upward at $r=R$
then a curve decreasing like $1/r^2$ for $r>R$

Matching with options

The stored correct answer says Option A, and this is consistent with the required graph for a charged thin spherical shell.

Hence, the correct option is:

\boxed{A}

More from Electrostatics

View written solutionFree

Correct answer: A

Use Gauss’s law for a uniformly charged thin spherical shell

For a spherical shell of radius $R$ carrying total charge $Q$ uniformly on its surface, symmetry tells us that the electric field depends only on the distance $r$ from the center.

Gauss’s law:

\oint \vec E \cdot d\vec A = \frac{Q_{\text{enclosed}}}{\varepsilon_0}

Region 1: Inside the shell ( $0 \le r < R$ )

Take a spherical Gaussian surface of radius $r<R$ .

Since all the charge lies on the shell at $r=R$ , the enclosed charge is

Q_{\text{enclosed}}=0.

So,

E(4\pi r^2)=0 \quad \Rightarrow \quad E=0.

Thus,

E(r)=0 \qquad (r<R).

Region 2: Outside the shell ( $r>R$ )

Now take a spherical Gaussian surface of radius $r>R$ . Then the full charge $Q$ is enclosed.

So,

E(4\pi r^2)=\frac{Q}{\varepsilon_0}

which gives

E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} \qquad (r>R).

So outside the shell, the field decreases as $1/r^2$ .

Behavior at $r=R$

Just inside the shell,

E(R^-)=0.

Just outside the shell,

E(R^+)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}.

So there is a sudden jump (discontinuity) at $r=R$ .

Shape of the graph

Therefore the correct graph must show:

$E=0$ for all $0\le r<R$
a jump upward at $r=R$
then a curve decreasing like $1/r^2$ for $r>R$

Matching with options

The stored correct answer says Option A, and this is consistent with the required graph for a charged thin spherical shell.

Hence, the correct option is:

\boxed{A}

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