Question

Two unit negative charges are placed on a straight line.
A positive charge q is placed exactly at the mid point
between these unit charges. If the system of these three
charges is in equilibrium, the value of q (in C) is​

Answer 1

Answer:

Now total force acting on (-1) at A is

F

net

=

r

2

k(−1)(−1)

+

(

2

r

)

2

k(−1)(q)

=0

r

2

1

=

r

2

4q

q=

4

1

=0.25C.

∴ Value of q for these three charges to be in equilibrium =0.25C

solution