Question

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

Solution

## $$\textbf{Part (a)}$$ $$\textbf{Step 1: Calculation of all electric forces}$$  $$\textbf{[Refer Fig. 1]}$$From Coulomb's Law-Electric force on $$Q$$ by $$q \quad (F_q) = \dfrac{KQq}{a^2}$$Electric force on $$Q$$ by $$Q\quad (F_Q) = \dfrac{KQ^2}{(\sqrt{2}a)^2} = \dfrac{KQ^2}{2a^2}$$$$\textbf{Step 2: Resultant of forces}$$    $$\textbf{[Refer Fig. 2]}$$Now, from fig. (2) we have to find the Resultant of all three forces.Resultant of both $$F_q$$ (Both are at $$90^o$$)               $$F' =\sqrt{F_q^2+ F_q^2}= \sqrt{2} \dfrac{KQq}{a^2}$$Now Resultant of $$F'$$ and $$F_Q$$  (Both are in same direction)               $$F_{net} = F' + F_Q$$$$=\sqrt{2} \dfrac{KQq}{a^2} + \dfrac{KQ^2}{2a^2}$$              $$\Rightarrow \boxed{F_{net} = \dfrac{KQ}{2a^2} (Q + 2\sqrt{2}q)}$$Direction of  $$F_{net}$$  will be same as direction of $$F_Q$$$$\textbf{Part b)}$$$$\textbf{Step 3: Potential energy of the system}$$  $$\textbf{[Refer Fig. 3]}$$There are $$6$$ No. of pairs possible in system as shown in figure.   So,  $$U_1 = \dfrac{KQq}{a}$$,     $$U_2=\dfrac{KQq}{a}$$      $$U_3 = \dfrac{KQq}{a}$$,     $$U_4=\dfrac{KQq}{a}$$      $$U_5 = \dfrac{KQ^2}{\sqrt{2}a}$$,     $$U_6=\dfrac{Kq^2}{\sqrt{2}a}$$Now, Net Potential Energy of system-      $$U= U_1 + U_2 + U_3 + U_4 + U_5 + U_6$$      $$\boxed{U = \dfrac{4KQq}{a} + \dfrac{KQ^2}{\sqrt{2}a} + \dfrac{Kq^2}{\sqrt{2}a}}$$ Physics

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