Manipal Entrance Test 2015 Physics Paper

Let the third charge - q is placed at a distance of x from the charge 2g. Then, the potential energy of the system is
U=k[

2q(8q)

r

−

(2q)(q)

x

−

(8q)(q)

r−x

]
Here, k=

1

4πε0

U is minimum, where

2

x

+

8

r−x

is maximum
Let

2

x

+

8

r−x

=y
For y to be maximum,

dy

dx

=0
−

2

x2

+

8

(r−x)2

=0
⇒

x

r−x

=√

2

8

=

1

2

⇒x=

r

3

But at x=

r

3

,

d2y

dx2

= positive i.e. at x=

r

3

,y is minimum or U is maximum. So, there cannot be any point between them where the potential energy is minimum.