Question

Two equal point changes, $Q=+\sqrt{2}\mu C$ are placed at each of the two opposite corners of a square and equal point charges q at each of the other two comers. The value of q, so that the resultant force on Q is zero is:

• A. $+0.5\mu C$
• B. $-0.5\mu C$
• C. $1\mu C$
• D. $-1\mu C$
• E. none of above

Answer 1

Answer: (B)

From Coulombs law $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{2a^2}$
$F_1$ and $F_2$ will be directed as shown, for this both q should be negative $F_1=F_2=\frac{1}{4\pi {\epsilon }_0}.\frac{Qq}{a^2}$ $F_{12}=\frac{1}{4\pi {\epsilon }_0}.\frac{\sqrt{2}Qq}{a^2}$ For equilibrium of Q, we have $F=-F_{12}$ $\frac{Q^2}{2a^2}=\frac{\sqrt{2}Qq}{a^2}$ $\Rightarrow$ $q=\frac{Q}{2\sqrt{2}}$ Given, $Q=+\sqrt{2}\mu C$ $\therefore$ $q=-\frac{\sqrt{2}\mu C}{2\sqrt{2}}=-0.5\mu C$