Question

A particle of charge q is placed at a distance r from the center of a uniformly charged spherical shell of radius R and charge Q, determine the electric force F experienced by the particle

Answer 1

The electric force F experienced by the particle of charge q depends on its position relative to the spherical shell:

1. If the particle is outside the spherical shell (r > R): The electric force is given by: $F = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r^2}$ The force is directed radially, being repulsive if Qq > 0 and attractive if Qq < 0.

2. If the particle is inside the spherical shell (r < R): The electric force is: $F = 0$

Answer 2

The electric force experienced by a particle of charge 'q' placed at a distance 'r' from the center of a uniformly charged spherical shell of radius 'R' and total charge 'Q' depends on whether the particle is located outside or inside the shell.

This problem is solved using the principles of electrostatics, specifically Gauss's Law, which simplifies the calculation of electric fields for symmetric charge distributions.

Case 1: Particle is outside the spherical shell (r > R)

When the point charge 'q' is located outside the uniformly charged spherical shell (i.e., its distance 'r' from the center is greater than the shell's radius 'R'), the electric field produced by the spherical shell at that point is identical to the electric field produced by a point charge 'Q' (the total charge of the shell) placed at the center of the shell.

The electric field ($E$) at a distance 'r' from the center is given by: $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$

The electric force ($F$) experienced by the particle of charge 'q' in this electric field is given by $F = qE$. Substituting the expression for E: $F = q \left( \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \right)$ $F = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r^2}$

The direction of this force is radial. It is repulsive if Q and q have the same sign (Qq > 0) and attractive if they have opposite signs (Qq < 0).

Case 2: Particle is inside the spherical shell (r < R)

When the point charge 'q' is located inside the uniformly charged spherical shell (i.e., its distance 'r' from the center is less than the shell's radius 'R'), the electric field inside the shell is zero. This is a direct consequence of Gauss's Law: if we draw a spherical Gaussian surface concentric with the shell and with radius 'r' (where r < R), the total charge enclosed within this Gaussian surface is zero, as all the charge 'Q' resides on the surface of the shell. Since the enclosed charge is zero, the electric flux through the Gaussian surface is zero, implying that the electric field inside the shell is zero.

The electric field ($E$) at a distance 'r' from the center (for r < R) is: $E = 0$

The electric force ($F$) experienced by the particle of charge 'q' in this zero electric field is: $F = qE = q(0) = 0$