Question

Three charges q, 2q and q are to be placed on a straight wire of length 10 m.
Determine the positions where the charges should to be placed so that the
electrostatic potential energy of the system is a minimum

Answer 1

Answer:

The potential difference is minimum when they are placed 5 cm apart.  

Explanation:

The total potential difference by a 3 charge system is given by  

V = kQ1*Q2/x + kQ2*Q3/y

Here Q1, Q2 & Q3 are charges places in a stright line, with Q2 in middle.

X – distance between Q1 & Q2.

Y – Distance between Q2 & Q3.

K – Constant.

In the given question, the potential difference will be minimum if charge ‘q’ is placed in the middle.  

So on a 10 cm line, charge ‘q’ is at one end, ‘2q’ is another end. ‘q’ charge is in middle.  

Let x is the distance between ‘q’ and ‘q’ charges. (10 – x) is distance between ‘q’ and ‘2q’.

V = kq*q/x + kq*2q/(10 -x)

Now potential difference to be minimum dV/dx = 0.

dV/dx = - kq^2/x^2 + 2kq^2/(10 – x)^2 = 0

Simplifying we get

(10 – x)^2 – x^2 = 0

100 – 20x = 0

x = 5.

The potential difference is minimum when they are placed 5 cm apart.