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Chapter 21: Problem 64

A point charge \(q\) is at the center of a spherical shell of radius \(R\) carrying charge \(2 q\) spread uniformly over its surface. Write expressions for the electric field strength at (a) \(\frac{1}{2} R\) and (b) \(2 R\).

Short Answer

Expert verified
The electric field strength at \( \frac{R}{2} \) is \( \frac{q}{\pi\epsilon_0 R^2} \) and at \(2 R\) it is \( \frac{5q}{16\pi\epsilon_0 R^2} \).

Step by step solution

01

Understanding the Setup

Consider a point charge \(q\) placed at the center of a spherical shell of radius \(R\). The shell contains a uniformly distributed charge \(2q\). Inside the shell, at \(\frac{R}{2}\), only the field due to the central charge is felt because the net field due to the distributed charge on the shell is zero inside the shell.
02

Calculating Electric Field at \(\frac{R}{2}\)

Electric field at a distance \(r\) from the center due to a point charge \(q\) is given by \(E = \frac{q}{4\pi\epsilon_0 r^2}\). Substituting \(r = \frac{R}{2}\) we get \(E = \frac{q}{\pi\epsilon_0 R^2}\).
03

Calculating Electric Field at \(2R\)

Outside the shell, both the field due to the central charge and the shell are felt. Using superposition, calculate the total electric field as the sum of the individual fields. The electric field due to the point charge is given by \(E_q = \frac{q}{4\pi\epsilon_0 (2R)^2} = \frac{q}{16\pi\epsilon_0 R^2}\) and the electric field due to the sphere is given by \(E_s = \frac{2q}{4\pi\epsilon_0 (2R)^2} = \frac{q}{4\pi\epsilon_0 R^2}\) , so total field \(E=\frac{q}{16\pi\epsilon_0 R^2} + \frac{q}{4\pi\epsilon_0 R^2} = \frac{5q}{16\pi\epsilon_0 R^2}\).
04

Final Answer

Therefore, the electric field strength at \( \frac{R}{2} \) is \( \frac{q}{\pi\epsilon_0 R^2} \) and at \(2 R\) it is \( \frac{5q}{16\pi\epsilon_0 R^2} \).

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Most popular questions from this chapter

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Charges \(+2 q\) and \(-q\) are near each other. Sketch some field lines for this charge distribution, using eight lines for a charge of magnitude \(q\).

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