Question

Two points R and S are equidistant from two charges \( +Q \) and \( -2Q \). The work done in moving a charge \( -Q \) from point R to S is:

• A. Zero
• B. \( - \frac{Q}{4 \pi \epsilon_0 d} \)
• C. \( \frac{Q}{4 \pi \epsilon_0 d} \)
• D. \( \frac{3Q}{4 \pi \epsilon_0 d} \)

Hint:

When moving a charge between two points where the electric potential is the same, no work is done since the potential difference \( \Delta V = 0 \).

Answer 1

The Correct Option is A

To determine the work done in moving a charge \( -Q \) from point R to S, we first need to understand the electric potential at these points due to the charges \( +Q \) and \( -2Q \).

Assume the point charges \( +Q \) and \( -2Q \) are located at points A and B, respectively, and R and S are equidistant from A and B. Let the distance from A to R (and S) be \( d_1 \) and from B to R (and S) be \( d_2 \).

The electric potential \( V \) due to a point charge is given by:

\(V = \frac{kQ}{r}\)

where \( k = \frac{1}{4 \pi \epsilon_0} \) is Coulomb's constant and \( r \) is the distance from the charge.

At point R, the potential \( V_R \) is the sum of potentials due to \( +Q \) and \( -2Q \):

\(V_R = \frac{kQ}{d_1} - \frac{2kQ}{d_2}\)

Similarly, at point S, the potential \( V_S \) is:

\(V_S = \frac{kQ}{d_1} - \frac{2kQ}{d_2}\)

Since R and S are equidistant from both charges, the distances \( d_1 \) and \( d_2 \) are the same for points R and S, making \( V_R = V_S \).

The work done \( W \) in moving a charge \( q \) between two points in an electric field is given by:

\(W = q(V_S - V_R)\)

Substituting the values, we have:

\(W = -Q(V_S - V_R)\)

\(W = -Q(\frac{kQ}{d_1} - \frac{2kQ}{d_2} - (\frac{kQ}{d_1} - \frac{2kQ}{d_2}))\)

This simplifies to:

\(W = -Q(0) = 0\)

Therefore, the work done in moving the charge from R to S is Zero. Thus, the correct answer is Zero.