Target Exam

CUET

Subject

Physics

Chapter

Electric Charges and Fields

Question:

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then the Q/q equals

Options:

$-2\sqrt{2}$

$-1$

$1$

$-\frac{1}{\sqrt{2}}$

Correct Answer:

$-2\sqrt{2}$

Explanation:

$ \text{ Force on Q due to charge q is } F_1 = \frac{kqQ}{a^2}$

$\text { a is length of side of square }$

$ \text{Force due to charges q on Q are perpendicular so resultant of two are }$

$F_{1R} = \frac{kqQ\sqrt 2}{a^2}$

$\text{Force due to charge Q } F_2 = \frac{KQ^2}{2a^2} $

$\text{Since resultant force is zero } F_{net} = 0$

$ \Rightarrow  \frac{kqQ\sqrt 2}{a^2} + \frac{KQ^2}{2a^2} =0$

$\Rightarrow q = -\frac{Q}{2\sqrt 2}$

$\Rightarrow \frac{Q}{q} = -2\sqrt 2$