Question

13.
In the given figure calculate the force on a charge Q placed at the centre of the circle of
radiusr.​

Answer 1

Given :

Radius of the given circle = 'r'

Charge at point A = -q

Charge at point B = +q

Charge at point C = +q

Charge at point D = -q

To  Find :

Force acting on the charge Q placed at the centre of this circle = ?

Solution :

Force acting on Q due to charge -q placed at A is :

=$\frac{K\times q \times Q}{r^2}$

Force acting on Q due to charge +q placed at B is :

=$\frac{K\times q \times Q}{r^2}$

Force acting on Q due to charge +q placed at C is :

=$\frac{K\times q \times Q}{r^2}$

Force acting on Q due to charge -q placed at D is :

=$\frac{K\times q \times Q}{r^2}$

Now the net force acting on Q will be given by the resultant of all the forces acting :

The diagrams for the direction of forces is shown in the attached fig .

$F_{net} = \sqrt{(\frac{2KqQ}{r^2}^2) + (\frac{2KqQ}{r^2}^2)}$

       =$\sqrt{8(\frac{KqQ}{r^2})^2}$

       =$\frac{2 \sqrt2 KqQ}{r^2}$

Hence , the net force acting on the charge Q is $\frac{2 \sqrt2 KqQ}{r^2}$ .

Answer 2

Where did 2 come from in 2kqQ²/r²??