Question

Four charges, $+q,+q,-q,-q$ are placed on the circumference of a circle of radius $r$. What is the force on the charge $+Q$ placed at the centre of the circle as shown in the figure?

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{\sqrt{8}Qq}{r^2}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{8Qq}{r^2}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{2Qq}{r^2}$

Answer 1

Answer: (B)

Force on charge $Q$ due to charge $q$ at $A$,
$F_1=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}$ along $AO$
Force on charge $Q$ due to charge $q$ at $B$,
$F_2=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}$ along $BO$
Force on charge $Q$ due to charge $-q$ at $C$,
$F_3=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}$ along $OC$
Force on charge $Q$ due to charge $-q$ at $D$,
$F_4=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}$ along $OD$
$\left|F_1\right|=\left|F_2\right|=\left|F_3\right|=\left|F_4\right|=F$
The resultant force on charge $Q$ is
$F_{net}=\sqrt{{\left(2F\right)}^2+{\left(2F\right)}^2+2\left(2F\right)\left(2F\right)cos90^{\circ }}$
$=F\sqrt{8}$
$=\frac{1}{4\pi {\epsilon }_0}\frac{\sqrt{8}Qq}{r^2}$