Question

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of them has charge −q. What is the magnitude of the electric force on the −q charge due to the three q charges?

Answer 1

Answer:
`F_(net)=(kq^2)/((L)^2)\left [ (1)/(2)+√(2)\right ]`

Step-by-step explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

`F_1=(kq(-q))/((L√(2))^2)`

where
`L√(2)=`Distance between the two charges

`F_1=-(kq^2)/(2L^2)`

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

`F_2=(kq(-q))/((L)^2)`

The magnitude of force by both the charge is same but at an angle of
`90^(\circ)`

thus combination of two forces at 2 and 3 will be

`F'=√(2)(kq^2)/(2L^2)`

Now it will add with force due to 1 charge

Thus net force will be

`F_(net)=(kq^2)/((L)^2)\left [ (1)/(2)+√(2)\right ]`