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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.
800575_fe4b600f43b040319e2f6f5b4cb20e86.png


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

      $$\boxed{U = \dfrac{4KQq}{a} + \dfrac{KQ^2}{\sqrt{2}a} + \dfrac{Kq^2}{\sqrt{2}a}}$$ 


2110600_800575_ans_b6403db25581478a80a335926a2072e4.png

Physics

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