Chapter 3, Problem 3.1P

(a)

To determine

The force on the point charge before the hemispheres are assembled around $Q_1$ .

Answer to Problem 3.1P

The force on a point charge $Q_2$ due to $Q_1$ before hemispheres are assembled around $Q_1$ is

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{R^2}{a}_r$

Explanation of Solution

Given:

The central charge $Q_1$ is located at the origin and a radius of the hemisphere is a while performing a Faraday concentric sphere experiment. In the second charge $Q_2$ , a point charge is located at distance $R$ from $Q_1$ , where $R>>a$ .

Calculation:

The force on the point charge $Q_2$ due to $Q_1$ without hemispheres being assembled around $Q_1$ is to be found. When no hemisphere is assembled around $Q_1$ , both the charges exist in free space with no other dielectric medium. $Q_2$ is located at a distance of R from $Q_1$ and since $R>>a$ , $Q_1$ can be considered as a point charge. Hence, in this case, the force on $Q_2$ due to $Q_1$ is that of force exerted by a point charge $Q_1$ on another point charge $Q_2$ .

The force on $Q_2$ due to $Q_1$ is equal to force between two-point charges in free space separated by distance R, given by

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{R^2}{a}_r$

The direction of the force is radially outward, the repulsive force from $Q_1$ to $Q_2$ as denoted by the unit vector $a_r$ .

Conclusion:

The force on $Q_2$ due to $Q_1$ before hemispheres are assembled around $Q_1$ is given by:

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{R^2}{a}_r$

(b)

To determine

The force on the point charge after the hemispheres are assembled but before they are discharged.

Answer to Problem 3.1P

Force on the point charge is $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{R^2}{a}_r$ .

Explanation of Solution

Given:

The central charge $Q_1$ is located at the origin and a radius of hemisphere is a, while performing a Faraday concentric sphere experiment. In a second charge $Q_2$ , a point charge is located at distance $R$ from $Q_1$ , where $R>>a$ .

Calculation:

In this case, when a hemisphere with radius a is assembled around $Q_1$ , a charge equal to $Q_1$ kept at the center is induced on the outer surface of the hemisphere due to the concept of displacement of charge. The charge is uniformly distributed over the outer surface of the hemisphere. Now since $R>>a$ , $Q_1$ can still be treated as a point charge and hence, the force between the charges $Q_1$ and $Q_2$ is still that of the force between two point charges.

The force on $Q_2$ due to $Q_1$ is equal to force between two-point charges in free space separated by distance R, given by:

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{R^2}{a}_r$

The direction of the force is radially outward, the repulsive force from $Q_1$ to $Q_2$ is denoted by the unit vector $a_r.$

Conclusion:

The force on $Q_2$ due to $Q_1$ after hemispheres are assembled around $Q_1$ and it is not discharged is given by:

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{R^2}{a}_r$

(c)

To determine

The force on the point charge after the hemispheres are assembled and after they are discharged

Answer to Problem 3.1P

The force on the point charge is zero after the hemispheres are assembled and discharged.

Explanation of Solution

Given:

The central charge $Q_1$ is located at the origin and a radius of the hemisphere is a while performing a Faraday concentric sphere experiment. In a second charge $Q_2$ , a point charge is located at distance $R$ from $Q_1$ , where $R>>a$ .

Calculation:

Since the outer surface of the hemisphere is grounded, the effective charge over the outer surface due to $Q_1$ is zero. This is equivalent to shielding the charge $Q_1$ . Thus, the only charge that exists $Q_2$ . Since the only charge is $Q_2$ , there is no force between two charges. So:

  $F=0$

Conclusion:

The force between the two charges when the hemispheres are discharged is $F=0$

(d)

To determine

The condition when charge $Q_2$ is moved toward the sphere assembly to the extent that the condition $R>>a$ is no longer valid.

Answer to Problem 3.1P

The force will become attractive.

Explanation of Solution

The central charge $Q_1$ is located at the origin and a radius of hemisphere is a while performing a Faraday concentric sphere experiment.

It is considered that $Q_1$ is a point charge in cases (a) to (c) since point charge $Q_2$ was kept at a distance $R>>a$ . But when the charge $Q_2$ is moved toward the sphere assembly such that $R\text{ << }a$ , $Q_2$ comes inside the sphere, and it induces a negative charge inside the sphere. Hence, the force between the two charges $Q_1$ and $Q_2$ becomes attractive which further increases as the charge moves more inside the sphere.