Point charges of equal magnitudes (25 nC) and opposite signs are placed on (diagonally) opposite corners of a 60-cm *80-cm rectangle.If point A is the corner of this rectangle nearest the positive charge and point B is located at the intersection of the diagonals of the rectangle, determine the potential difference, VB - VA.

Question

Quizplus · Accepted Answer

The potential difference between two points in an electric field is given by the negative of the work done per unit charge in moving a test charge from one point to the other. In this case, the work done in moving a positive test charge from point A to point B is equal to the work done in moving the charge from point A to infinity minus the work done in moving the charge from infinity to point B. The work done in moving the charge from point A to infinity is given by:$$W_{A\rightarrow\infty}=\frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_1}+\frac{1}{4\pi\varepsilon_0}\frac{q_2}{r_2}$$where q1 and q2 are the charges at points A and B, respectively, r1 is the distance from point A to the charge at point B, and r2 is the distance from point A to infinity. The work done in moving the charge from infinity to point B is given by:$$W_{\infty\rightarrow B}=\frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_3}+\frac{1}{4\pi\varepsilon_0}\frac{q_2}{r_4}$$where r3 is the distance from point B to the charge at point A, and r4 is the distance from point B to infinity. The potential difference between points A and B is then given by:$$V_B-V_A=W_{A\rightarrow\infty}-W_{\infty\rightarrow B}$$$$=\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}-\frac{q_1}{r_3}-\frac{q_2}{r_4}\right]$$$$=\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1(r_3-r_1)}{r_1r_3}+\frac{q_2(r_4-r_2)}{r_2r_4}\right]$$Since the charges are equal in magnitude but opposite in sign, we have q1 = -q2. Substituting this into the above equation, we get:$$V_B-V_A=\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1(r_3-r_1)}{r_1r_3}-\frac{q_1(r_4-r_2)}{r_2r_4}\right]$$$$=\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1(r_3-r_1-r_4+r_2)}{r_1r_2r_3r_4}\right]$$$$=\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1(0)}{r_1r_2r_3r_4}\right]$$$$=0$$Therefore, the potential difference between points A and B is zero.

Point Charges of Equal Magnitudes (25 NC) and Opposite Signs