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  • Two opposite corners of a square carry the charge each and the remaining two carry a charge <img alt="\mathrm{Q}" src="https:

Two opposite corners of a square carry the charge U each and the remaining two carry a charge \mathrm{Q} each. The resultant force on \mathrm{Q}  is found to be zero. How are \mathrm{U} and \mathrm{Q} related?

Option: 1

Q=-2 \sqrt{2} U


Option: 2

Q=-\frac{2}{\sqrt{2}} U


Option: 3

U=-2 \sqrt{2} Q


Option: 4

U=-\frac{2}{\sqrt{2}} Q


Answers (1)

best_answer

Let each side of the square = x.

Then, diagonal, 

\begin{gathered} \sqrt{x^2+x^2}=x \sqrt{2} \\ F_1=F_2=\frac{U Q}{4 \pi \epsilon_o x^2} \\ F_3=\frac{Q Q}{4 \pi \epsilon_o(x \sqrt{2})^2}=\frac{Q^2}{2.4 \pi \epsilon_o x^2} \end{gathered}

As both forces,  F_1 and F_2  are perpendicular to each other, resultant force =

\begin{aligned} & \quad F=\sqrt{\left(F_1\right)^2+\left(F_2\right)^2}=\sqrt{\left(F_1\right)^2+\left(F_1\right)^2} \\ & F=F_1 \sqrt{2} \end{aligned}

Net force on Q is zero, hence;

\begin{aligned} & F_1 \sqrt{2}=-F_3 \\ & \frac{U Q \sqrt{2}}{4 \pi \epsilon_o x^2}=\frac{-Q^2}{2 \pi \times 4 \epsilon_o x^2} \\ & Q=-2 \sqrt{2} U \end{aligned}

Posted by

Ritika Jonwal

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