Question

Four point charges $Q,q,Q$ and $q$ are placed at the corners of a square of side ${}^{\prime }{a}^{\prime }$ as shown in the figure.
Find the
(a) resultant electric force on a charge $Q$, and
(b) potential energy of this system.

Answer 1

$\text{Part (a)}$

$\text{Step 1: Calculation of all electric forces}$ $\text{[Refer Fig. 1]}$

From Coulomb's Law-

Electric force on $Q$ by $q\left(F_q\right)=\frac{KQq}{a^2}$

Electric force on $Q$ by $Q\left(F_Q\right)=\frac{K{Q}^2}{(\sqrt{2}a{)}^2}=\frac{K{Q}^2}{2a^2}$

$\text{Step 2: Resultant of forces}$ $\text{[Refer Fig. 2]}$

Now, from fig. (2) we have to find the Resultant of all three forces.

Resultant of both $F_q$ (Both are at ${90}^o$)

$F^{\prime }=\sqrt{F_q^2+F_q^2}=\sqrt{2}\frac{KQq}{a^2}$

Now Resultant of $F^{\prime }$ and $F_Q$ (Both are in same direction)

$F_{net}=F^{\prime }+F_Q$$=\sqrt{2}\frac{KQq}{a^2}+\frac{K{Q}^2}{2a^2}$

$\Rightarrow F_{net}=\frac{KQ}{2a^2}(Q+2\sqrt{2}q)$

Direction of $F_{net}$ will be same as direction of $F_Q$

$\text{Part b)}$

$\text{Step 3: Potential energy of the system}$ $\text{[Refer Fig. 3]}$

There are $6$ No. of pairs possible in system as shown in figure.

So, $U_1=\frac{KQq}{a}$, $U_2=\frac{KQq}{a}$

$U_3=\frac{KQq}{a}$, $U_4=\frac{KQq}{a}$

$U_5=\frac{K{Q}^2}{\sqrt{2}a}$, $U_6=\frac{K{q}^2}{\sqrt{2}a}$

Now, Net Potential Energy of system-

$U=U_1+U_2+U_3+U_4+U_5+U_6$

$U=\frac{4KQq}{a}+\frac{K{Q}^2}{\sqrt{2}a}+\frac{K{q}^2}{\sqrt{2}a}$