A charge q is placed at every corner of a square having side a. how much charge must be placed at its centre to bring the system in equilibrium?

 Let the side length of the square be a.

In △ACD,

⇒AD2 + CD2 = AC2

⇒ a2 + a2 =AC2 (AC=√2a)

However, OC = AC/2 = a/√2

Because of the diagonally opposing charges cancelling each other, the force on the centre charge will be equal and opposite according to symmetry. As a result, regardless of the charge q value, the net force on the centre charge will always be zero. To discover the value of q, we must first examine the equilibrium of any single charge at the corner.

Force on charge at C due to B,F1 = KQ2/a2

Force on charge at C due to D,F2 = KQ2/a2

Force on charge at C due to A,F4=KQ2/AC2=KQ2/2a2

Force on charge at C due to q at centre,F3=KqQ/OC2= 2KqQ/a2

Resultant of F1 and F2 (Along OC, by symmetry) = √F12+F2= √2F1

Since (∣F1∣=∣F2∣)
Also, F3 & F4 are along OC, Therefore, magnitudes of sum of these forces should be zero.
⇒ ∣F3∣ + √2∣F1∣ + ∣F4∣ = 0
⇒ 2KqQ/a2 +√2 KQ2/a2 +KQ2/2a2 =0
⇒ 2q = −(√2+Q/2)
⇒ q = −Q/4(1+2/2)

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