Question

Four point charges $q_A=2\mu C,q_B=-5\mu C,q_C=2\mu C,and{q}_D=-5\mu C$ are located at the corners of a square ABCD of side 10cm. What is the force on a charge of $1\mu C$ placed at the centre of the square?

Answer 1

Charge placed at the centre is in the influence field of four charges located at the corners of the square. Therefore, we can find force acting on charge placed at the centre using superposition principle. Use the law of vectors to find the net resultant force because force is a vector quantity.

Let the centre of the square is at O. The charge placed on the centre is $1\mu C$.
AB=BC=CD=DA=10cm
AC= $\sqrt{2}\times 10=10\sqrt{2}$ cm
$AC=BD=10\sqrt{2}cmAO=BO=CO=DO=\frac{10\sqrt{2}}{2}=5\sqrt{2}cm$



Let the force on charge $1\mu C$ due to $q_{A}is{F}_A$ which away from both charges $q_{A}andq$ (because both charges are positive in nature, so will repel each other).
The force in charge $1\mu C$ due to $q_c$ is $F_c$ which away from both $q_c$ and q (as they both are positive in nature, so will repel each other).
The force on charge $1\mu C$ due to $q_B$ is $F_B$ which is towards $q_B$ (because $q_B$ is negatively charged and q is positively charged, so will attract each other).
The force on charge $1\mu C$ due to $q_D$ is $F_D$ which is towards $q_D$ (because $q_D$ is negatively charged and q is positively charged, so will attract each other).

Force between q and $q_A$
$F_A=\frac{1}{4\pi {\epsilon }_0}.\frac{q{q}_A}{(OA{)}^2}=\frac{9\times {10}^9\times 1\times {10}^{-6}\times 2\times {10}^{-6}}{(5\sqrt{2}\times {10}^{-2}{)}^2}=\frac{9\times 2\times {10}^{-3}}{25\times 2\times {10}^{-4}}{F}_A=\frac{90}{25}=\frac{18}{5}=3.6N\left(directiontowardsOtoC\right)$
Force between q and $q_C$
$F_C=\frac{1}{4\pi {\epsilon }_0}.\frac{q{q}_C}{(OC{)}^2}=\frac{9\times {10}^9\times 1\times {10}^{-6}\times 2\times {10}^{-6}}{(5\sqrt{2}\times {10}^{-2}{)}^2}=\frac{9\times 2\times {10}^{-3}}{25\times 2\times {10}^{-4}}{F}_C=\frac{90}{25}=\frac{18}{5}=3.6N\left(directiontowardsOtoA\right)$
Here, we observe that $F_A$ and $F_C$ are of same magnitude and opposite in direction. So, the resultant force of $F_A$ and $F_C$ is zero.
Force between q and $q_B$
$F_B=\frac{1}{4\pi {\epsilon }_0}.\frac{q{q}_B}{(OB{)}^2}=\frac{9\times {10}^9\times 1\times {10}^{-6}\times 5\times {10}^{-6}}{(5\sqrt{2}\times {10}^{-2}{)}^2}=\frac{9\times 5\times {10}^{-3}}{25\times 2\times {10}^{-4}}{F}_B=9N\left(directiontowardsOtoB\right)$
Force between q and $q_D$
$F_D=\frac{1}{4\pi {\epsilon }_0}.\frac{q{q}_D}{(OD{)}^2}=\frac{9\times {10}^9\times 1\times {10}^{-6}\times 5\times {10}^{-6}}{(5\sqrt{2}\times {10}^{-2}{)}^2}=\frac{9\times 5\times {10}^{-3}}{25\times 2\times {10}^{-4}}{F}_D=9N\left(directiontowardsOtoD\right)$

Here, we observe that $F_B$ and $F_D$ are of some magnitude and opposite in direction. So, the resultant force of $F_D$ and $F_B$ is zero.
Thus, the net resultant force on $1\mu C$ (placed at O) is zero as all the forces balance each other.