Question

Four charges equal to Q are placed at four corners of a square and a charge q placed at the centre of the square are in equilibrium. Find the value of q.​

Answer 1

☆●Given:-

let the side of the square be a

☆To prove:-

Find the value of q

☆Proof:-

In triangle ACD

=AD^2+CD^2=AC^2

=a^2+a^2=AC^2

AC= root 2 a .

In triangle DOC,

=OD^2+OC^2=CD^2

OC^2+OC^2=a^2

OC=a/root2.

For the system to be equilibrium net force acting at point C must be zero.

F3=F1+F2+F4(Given |F1|=|F2|)

|F3|=root2|F1|+|F4|

2kqQ= root2 KQ^2 +KQ^2

a^2 a^2 a^2

2q=root2Q+Q/2

q=Q/4 (1+2root2).

Hope it helps u mate.

Thank you.

Mark it as BRAINLIEAST please i request.

Answer 2

Four charges equal to Q are placed at four corners of a square and a charge q placed at the center of the square are in equilibrium i.e the net force on the charge q  placed at the center is 0.

As a result, the net force on all the charges( corner as well as the center) will be zero.

Let, the distance between the 2 charges or the sides of the square be r.

First, we need to draw the net force acting upon charge Q

( Refer to the attachment )

Net force acting on Q,

$\to\sf{F_{net} + F_b +F_o=0}$

$\to\sf{F_{net} + F_b =- F_o }$

$\to\sf{\sqrt{2}F + F_b =- F_o }$

$\to\sf{\sqrt{2}\dfrac{KQ^2}{r^2} +\dfrac{KQ^2}{2r^2} =-\dfrac{2KQq}{r^2} }$

$\sf{Canceling \:out \:common \: terms \:from \: both \: the \: sides \: we \:get,}$

$\to\sf{\sqrt{2}Q^2 +\dfrac{Q^2}{2} =- 2Qq }$

$\to\sf{\dfrac{2\sqrt{2}Q^2 +Q^2}{2} = - 2Qq }$

$\to\sf{\dfrac{2\sqrt{2}Q^2 +Q^2}{4Q} =- q }$

$\to\sf{- q = \dfrac{2\sqrt{2}Q^2}{4Q} +\dfrac{Q^2}{4Q} }$

$\to\boxed{\sf{q =-\dfrac{2\sqrt{2}Q+Q}{4}}}$