question_answer

An electric charge\[{{y}^{2}}={{x}^{2}}-2xy\frac{dy}{dx}\]is placed at the origin (0, 0) of xy-coordinate system. Two points A and B are situated at\[{{p}^{2}}+{{q}^{2}}=1,\]and (2, 0) respectively. The potential difference between the points A and B will be          [AIEEE 2007]

A) 9V

B) zero

C) 2V

D) 4.5V

Correct Answer: B

Solution :

[b] The potential at any point due to system of charges is the sum of potentials due to individual charges that does not be altered while the changes as some of charges are interchanged. Potential at A due to charge at O.

\[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{({{10}^{-3}})}{OA}\]

\[12!/3!{{(4!)}^{3}}\]

potential at B due to charge at O

\[12!/3!{{(3!)}^{4}}\]

\[12!/{{(4!)}^{3}}\]

Distance formula is used to determine (OB).

So, \[12!/{{(3!)}^{4}}\]