Consider the force F on a charge q due to a uniformly charged spherical shell of radius R carrying c

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

According to Gauss's Law, the electric flux through any closed surface is proportional to the enclosed charge. If you draw a spherical Gaussian surface inside the hollow shell, it encloses zero charge. Due to spherical symmetry, the electric field itself must be zero everywhere inside.

Q: Does the shell theorem apply if the charge is not uniformly distributed?

No. The shell theorem relies heavily on perfect spherical symmetry. If the charge is concentrated more on one side, the electric field inside will no longer be zero, and the field outside will not behave exactly like a point charge at the center.

Q: What happens exactly at the surface where $r = R$?

At the exact mathematical boundary $r = R$, the electric field is technically discontinuous. However, for practical purposes in physics, the field just outside the surface is taken as $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2}$.

Q: Why can we treat the shell as a point charge for outside points?

This is a beautiful consequence of the inverse-square nature of Coulomb's law combined with spherical geometry. The contributions from the parts of the shell closer to the point exactly balance the contributions from the parts farther away, mathematically reducing to a single point charge at the center.

Q: Is the force attractive or repulsive?

The formula gives the magnitude of the force. Whether it is attractive or repulsive depends on the signs of $Q$ and $q$. If they have the same sign, it's repulsive; if opposite, it's attractive.

Question

Consider the force $F$ on a charge $q$ due to a uniformly charged spherical shell of radius $R$ carrying charge $Q$ distributed uniformly over it. Which one of the following statement is true for $F$, if $q$ is placed at distance $r$ from the centre of the shell ? Options: (A) $F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{R^2}$ (for $r < R$) (B) $\frac{1}{4\pi\varepsilon_0} \frac{Qq}{R^2} > F > 0$ (for $r < R$) (C) $F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2}$ (for $r > R$) (D) $F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{R^2}$ (for all $r$)

Sigma Percentile · Accepted Answer

This problem is a classic application of **Gauss's Law** and the **Shell Theorem**, which beautifully simplifies the electrostatic interactions of spherically symmetric charge distributions. ### Analyzing the Setup We are given a spherical shell of radius $R$ that carries a uniformly distributed charge $Q$. We need to determine the electrostatic force $F$ experienced by a test charge $q$ placed at a distance $r$ from the center of this shell. To find the force, we rely on the fundamental relationship between force and electric field: $$F = qE$$ Our primary objective is to find the electric field $E$ produced by the shell at the location of the charge $q$. Because the shell is spherically symmetric, we must consider two distinct regions: inside the shell ($r < R$) and outside the shell ($r > R$). ### Case 1: Inside the Shell ($r < R$) Imagine placing the charge $q$ anywhere inside the hollow region of the shell. According to Gauss's Law, if we draw a spherical Gaussian surface of radius $r < R$ concentric with the shell, the charge enclosed by this surface is exactly zero. Because the enclosed charge is zero and the setup is perfectly symmetric, the electric field must be zero everywhere inside the shell: $$E_{ ext{inside}} = 0$$ Consequently, the force on any charge placed inside the shell is also zero: $$F = q(0) = 0$$ This immediately tells us that options (a) and (b) are incorrect, as they suggest a non-zero force inside the shell. ### Case 2: Outside the Shell ($r > R$) Now, let's move the charge $q$ to a point outside the shell. If we draw a Gaussian surface of radius $r > R$, it encloses the entire charge $Q$ of the shell. Gauss's Law tells us that for any point outside a spherically symmetric charge distribution, the electric field is identical to that of a point charge $Q$ located exactly at the center. $$E_{ ext{outside}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$$ Multiplying this electric field by our test charge $q$, we obtain the force: $$F = qE = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2}$$ ### Final Conclusion Comparing our derived expressions with the given options, we find that option (c) perfectly matches our result for the region outside the shell. The force follows the inverse-square law, treating the entire shell as a point charge at its center.

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Visualized Solution

$The Setup$

$Electric Field of a Shell$

$Case 1: Inside the Shell (r < R)$

$Force Inside (r < R)$

$Case 2: Outside the Shell (r > R)$

$Force Outside (r > R)$

$Conclusion$

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

Analyzing the Setup We are given a spherical shell of radius $R$ that carries a uniformly distributed charge $Q$

Case 1

Case 2

Final Conclusion Comparing our derived expressions with the given options, we find that option (c) perfectly matches our result for the region outside the shell

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field of a Shell

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Outside the Shell (r>R)

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

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