question_answer

A point charge Q is placed at point O as shown in figure. Is the potential difference \[{{V}_{A}}-{{V}_{B}}\] positive, negative or zero, if Q is (i) positive, (ii) negative?

Answer:

            \[{{V}_{A}}-{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{OA}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{OB}=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{OA}-\frac{1}{OB} \right]\] From the diagram the distance OA < OB. \[\therefore \frac{1}{OA}-\frac{1}{OB}\] is positive.

(i) If Q is positive, then \[{{V}_{A}}-{{V}_{B}}\] is positive.

(ii) If 0 is negative, then  \[{{V}_{A}}-{{V}_{B}}\] is negative.