Question

as a result, the net electric force experienced by each negatively charged particle is reduced to F/2. The value of q is

Answer 1

The value of q is
`(Q)/(8)`

Step-by-step explanation:

Given that,

Each charge = -Q

Distance between charges = L

Reduced force =
`(F)/(2)`

Suppose, Two particles each with a charge -Q are fixed a distance L apart as shown above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles.

We know that,

The force on each end is

`F=(kQ^2)/(L^2)`...(I)

If the charge q is placed at mid point then

The force on each end charge is

`(F)/(2)=F+F'`....(II)

We need to calculate the value of q

Using equation (II)

`(F)/(2)=F+F'`

Put the value of F into the formula

`((kQ^2)/(L^2))/(2)=k(Q^2)/(L^2)+k(q*(-Q))/(((L)/(2))^2)`

`(kq(-Q))/(((L)/(2))^2)=-(kQ^2)/(2L^2)`

`(q)/((1)/(4))=(Q)/(2)`

`q=(Q)/(8)`

Hence, The value of q is
`(Q)/(8)`