Question

two charges are kept 30 cm apart. Where should another charge be placed so that it will be in equillibrium

Answer 1

Explanation:

Two point charges +9q and +9.

Kept apart =16cm=16×10

−2

m

Two cases will be needed.

Given charges, +q, +9q. let, the third charge q be placed at a distance x from q and hence its distance from 9q and (16−x)cm for the system to remain in equilibrium F

1

=F

2

Therefore,

F

1

=

x

2

K(q×Q)

F

2

=

(16−x)

2

K(9q×Q)

Equating F

1

and F

2

we get,

x

2

9

=

(16−x)

2

1

F

1

=F

2

or,

x

2

K(q×Q)

=

(16−x)

2

K(9q×Q)

or,

x

2

1

=

(16−x)

2

9

or,

x

1

=

(16−x)

9

or, 16−x=3x

or, 4x=16

or, x=4

∴ (16−4)cm=12cm from +9q.