Target Exam

CUET

Subject

Physics

Chapter

Electric Charges and Fields

Question:

Two point charges $q_1 =+2C$ and $q_2 =-1C$ are separated by a distance $d$. The position on the line joining the two charges where a third charge $q = + 1C$ will be in equilibrium is at a distance

Options:

$d/\sqrt{2}$ from $q_1$ between $q_1$ & $q_2$

$d/\sqrt{2}$ from $q_1$ away from $q_2$

$d/\sqrt{2} – 1$ from $q_2$ between $q_1$ & $q_2$

$d/\sqrt{2}-1$ from $q_2$ away $q_1$

Correct Answer:

$d/\sqrt{2}-1$ from $q_2$ away $q_1$

Explanation:

Let electric field is zero at a distane x from q2 charge away from q1.

$\frac{K.q_2q}{4\pi \epsilon_0 x^2} = \frac{K q_1q}{4\pi \epsilon_0 (x+d)^2}$

$\Rightarrow (x+d)^2 = 2 x^2$

$\Rightarrow (x+d) = \pm \sqrt2 x$

$\Rightarrow x = \frac{d}{\sqrt2 -1} $