Question

Four positive charges $(2\sqrt{2}-1)Q$ are arranged at the four corners of a square. Another charge $q$ is placed at the centre of the square. For what value of $q$ , is the resulting force acting on each corner charge be zero ?

• A. $-Q$
• B. $-(\sqrt{2}+1)Q$
• C. $\frac{-4Q}{7}$
• D. $\frac{-7Q}{4}$

Answer 1

The correct option is D $\frac{-7Q}{4}$

Given,

$Q_1=(2\sqrt{2}-1)Q$

Let,
$|F_1|\to$ Magnitude of force between charges $Q_1$ and $Q_1$, kept at a distance ${}^{\prime }{a}^{\prime }$ at two adjacent cornes. (direction shown in figure)

$|F_2|\to$ Magnitude of force between $Q_1$ and $Q_1$ kept at a distance ${}^{\prime }\sqrt{2}{a}^{\prime }$ at diagonally opposite corners. (direction shown in figure)

$|F_3|\to$ Magnitude of force between $Q_1$ and $q$ at a distance $\frac{a}{\sqrt{2}}$

For $F_3$ to be in the shown direction, $q$ and $Q_1$ should have opposite signs.

For net force to be zero on $Q_1$ placed at $P$.
$|F_3|=|\text{Resultant of}{F}_1,F_2\text{and}{F}_1|$

Since $F_2$ makes an angle of ${45}^{\circ }$ with both the edges of the square, we can resolve both the $F_1$ forces along the diagonal of the square.

We see that the components of $F_1$ along the normal to the diagonal in the plane is equal in magnitude and opposite in direction and hence, get cancelled out.

The components along the diagonal get added up with $F_2$ and are equal in magnitude and opposite in direction to $F_3$ to get an equilibrium position at $P$.

$\Rightarrow |F_3|=\sqrt{2}|F_1|+|F_2|....\left(1\right)$

Now from coulomb's law of electrostatic forces,

$|F_3|=\frac{K{Q}_{1}q}{{\left(\frac{a}{\sqrt{2}}\right)}^2}$

$|F_1|=\frac{K{Q}_1{Q}_1}{a^2}$

$|F_2|=\frac{K{Q}_1{Q}_1}{(\sqrt{2}a{)}^2}$

Substituting the values in the equation$\left(1\right)$, we get,

$\frac{K{Q}_{1}q}{{\left(\frac{a}{\sqrt{2}}\right)}^2}=\sqrt{2}\left(\frac{K{Q}_1{Q}_1}{a^2}\right)+\frac{K{Q}_1{Q}_1}{(\sqrt{2}a{)}^2}$

$\Rightarrow |q|=\frac{Q_1}{2}(\sqrt{2}+\frac{1}{2})$

$\Rightarrow |q|=\left(\frac{2\sqrt{2}+1}{4}\right)Q_1$

On substituting, $Q_1=(2\sqrt{2}-1)Q$,

$\Rightarrow |q|=\left(\frac{2\sqrt{2}+1}{4}\right)(2\sqrt{2}-1)Q$

$\Rightarrow |q|=\frac{7}{4}Q$

Since, the electrostatic force is taken in the direction of $q$ and $Q$ is positive, we can say that $q$ is negative.

$\Rightarrow q=-\frac{7}{4}Q$

Hence, option (a) is correct answer.