Question

A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A point charge q is fixed to each of the remaining corners. The net force acting on either of the charges q is zero. Find the magnitude and algebraic sign of q.

Answer 1

Answer:

The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

$F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The electric force at opposite corner

$F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The net force is

$F=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}$

The electric force at second corner

$F_{2}=\dfrac{-kq^2}{2r^2}$

The net force acting on either of the charges is zero.

So,  $F=F'$

$\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}$

$\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}$

$q=14\sqrt{2}\ \mu C$

Hence, The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$