# 6Two free point charges +q and +4q are a distance apart.A third charge is placed so that the entire system is in equilibrium. Find the location, magnitude and sign of the third charge?

Arun
25750 Points
5 years ago

The given charges q and 4q are free and they repel each other. Therefore , the third change should be such that it holds both the charges. Therefore, the third charge must be negative. Also, it should be placed in between the two charges.

Let the distance between q and 4q be a. We place -q1 charge at distance x from q.

For equilibrium of the system, the potential energy U of the system must be minimum.

Now, U=- kqq1/x +k4q^2/a - k4qq1/(a-x)….(1)

For, U to be minimum,dU/dx=0. Therefore ,

dU/dx=0=kqq1/x^2-k4qq1/(a-x)^2. Then,

1/x^2= 4/(a-x)^2

(a-x)^2=4x^2

(a-x)=+_2x

a-x=2x or

x=a/3……….(2).

With this value of x, the potential energy of the system will be, from equation (1),

-kqq1/(a/3)+4kq^2/a-4kqq1/(2/3a)=U…………..(3)

Now, if we want to find potential energy, we must know the value of q1.

Now, q1 can be found by equating force on q OR 4q to zero.

Let us consider force on q.

-9kqq1/a^2+kq(4q)/a^2=0. Then,

q1=(4/9)q in value.

Substituting this value of q1 in equation (3), we can find U.

U will come out to be -(55/6)kq^2/a.