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Chapter 13: Problem 82

Two concentric spheres are of radii \(r_{1}\) and \(r_{2}\). The outer sphere is given a charge \(q\). The charge \(q^{\prime}\) on the inner sphere will be (inner sphere is grounded) (A) \(q\) (B) \(-q\) (C) \(-q \frac{r_{1}}{r_{2}}\) (D) Zero

Short Answer

Expert verified
The charge on the inner grounded sphere, \(q'\), is given by: \(-q \frac{r_{1}}{r_{2}}\).

Step by step solution

01

Understand the problem

We have two concentric spheres of radii \(r_{1}\) and \(r_{2}\) where outer sphere has a charge \(q\) and the inner sphere is grounded. We need to find the charge on the inner sphere (\(q'\)).
02

Set up Gauss's Law

We will apply Gauss's Law: \[\oint_{S} \vec{E} \cdot \vec{dA} = \frac{q_{enclosed}}{\epsilon_0}\] Here, \(\vec{E}\) is the electric field due to the charged outer sphere, \(\vec{dA}\) is the area element vector and \(\epsilon_0\) is the electric constant. We will calculate the surrounding electric field at a distance \(r\) from the center of the spheres. Since the electric field is radial and the area element vector is also radial, \(\vec{E} \cdot \vec{dA} = E \cdot dA\).
03

Calculate the electric field

Consider a Gaussian surface, a sphere with a radius (\(r_{1} < r < r_{2}\)) between the two spheres. Applying Gauss's law: \[E\oint_S dA = \frac{q}{\epsilon_0}\] Since \(E\) is constant on the Gaussian surface: \[E \cdot 4\pi r^2 = \frac{q}{\epsilon_0}\] Solving for \(E\): \[E = \frac{q}{4\pi\epsilon_0 r^2}\]
04

Calculate the potential with respect to ground

To find potential, we integrate the electric field across the distance from the outer sphere to the inner sphere: \[\Delta V = -\int_{r_2}^{r_1} E \cdot dr\] Use the electric field from step 3: \[\Delta V = -\int_{r_2}^{r_1} \frac{q}{4\pi\epsilon_0 r^2} dr\] After integration, we get: \[\Delta V = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)\] But since the inner sphere is grounded, its potential is zero: \[\Delta V = 0\] So: \[0 = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)\]
05

Calculate the charge on the inner sphere

We can solve for q' with the obtained relation in the potential: \[0 = q - q'\left(\frac{r_2}{r_1}\right)\] Solving for \(q'\): \[q' = q\frac{r_1}{r_2}\] Thus, the correct answer is (C) \(-q \frac{r_{1}}{r_{2}}\).

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Most popular questions from this chapter

Two identical particles each having charge \(q\) are very far apart. They are given velocity \(v_{0}\) parallel to each other such that initial perpendicular separation between them is \(d\). If the subsequent minimum separation between them is \(2 d\), find the initial velocity \(v_{0}\) and the loss in their total kinetic energies.

There is an electric field \(E\) in \(x\)-direction. If work done in moving a charge \(0.2 \mathrm{C}\) through a distance of \(2 \mathrm{~m}\) along a line making an angle of \(60^{\circ}\) with \(x\)-axis is 4.0 J. The value of \(E\) is (A) \(\sqrt{3} \mathrm{~N} / \mathrm{C}\) (B) \(4 \mathrm{~N} / \mathrm{C}\) (C) \(5 \mathrm{~N} / \mathrm{C}\) (D) \(20 \mathrm{~N} / \mathrm{C}\)

The electric field at the origin is along the positive \(X\)-axis. A small circle is drawn with the centre at the origin cutting the axes at points \(A, B, C\), and \(D\) having coordinates \((a, 0)(0, a),(-a, 0),(0,-a)\), respectively. Out of the points on the periphery of the circle, the potential is minimum at (A) \(A\) (B) \(B\) (C) \(C\) (D) \(D\)

Two point charges \(+4 q\) and \(+q\) are placed \(30 \mathrm{~cm}\) apart. At what point on the line joining them is the electric field zero? (A) \(15 \mathrm{~cm}\) from charge \(4 q\) (B) \(20 \mathrm{~cm}\) from charge \(4 q\) (C) \(7.5 \mathrm{~cm}\) from charge \(q\) (D) \(5 \mathrm{~cm}\) from charge \(q\)

A point charge \(q\) is placed at origin. Let \(\vec{E}_{A}, \vec{E}_{B}\) and \(\vec{E}_{C}\) be the electric field at three points \(A(1,2,3)\), \(B(1,1,-1)\), and \(C(2,2,2)\) due to charge \(q\). Then (A) \(\vec{E}_{A} \perp \vec{E}_{B}\) (B) \(\vec{E}_{A} \| \vec{E}_{B}\) (C) \(\left|\vec{E}_{B}\right|=4\left|\vec{E}_{C}\right|\) (D) \(\left|\vec{E}_{B}\right|=16\left|\vec{E}_{C}\right|\)
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