Three charges, each equal to +q, are placed at the corners of an equilateral triangle ABC of side 'a' as shown in the figure. P and R are the mid-points of BC and CA. The work done in taking a charge Q from point P to R is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Zero
The electrostatic force is conservative, so the work done in moving Q depends only on the potentials at the two ends: \( W = Q\,(V_R - V_P) \).
P is the mid-point of BC, so it is \( a/2 \) from B and from C and \( \sqrt{3}\,a/2 \) (the height of the triangle) from A. R is the mid-point of CA, so it is \( a/2 \) from C and from A and \( \sqrt{3}\,a/2 \) from B. With \( k = 1/4\pi\varepsilon_0 \):
\( \displaystyle V_P = kq\left(\frac{2}{a} + \frac{2}{a} + \frac{2}{\sqrt{3}\,a}\right) \)
\( \displaystyle V_R = kq\left(\frac{2}{a} + \frac{2}{a} + \frac{2}{\sqrt{3}\,a}\right) \)
\( \displaystyle V_P = V_R \;\Rightarrow\; W = Q\,(V_R - V_P) = 0 \)
P and R are placed symmetrically with respect to the three equal charges, so no calculation is really needed.