CBSE Class 12 · 2005 · All India · ElectrostaticsShort Answer

A charge

q

is placed at the centre of the line joining two equal charges

Q

. Show that the system of three charges will be in equilibrium if

q = - \frac{Q}{4}

Answer: $q = - \frac{Q}{4}$ (a negative charge of one-quarter the magnitude).

Solution

Let the two charges

Q

be separated by

2 x

, with

q

at the mid-point.
By symmetry the force on

q

is zero, so

q

is already in equilibrium.
For an outer charge

Q

, the repulsion by the other

Q

and the force due to

q

must cancel:

\frac{1}{4 π ε _{0}} \frac{Q ^{2}}{( 2 x ) ^{2}} + \frac{1}{4 π ε _{0}} \frac{Qq}{x ^{2}} = 0

\frac{Qq}{x ^{2}} = - \frac{Q ^{2}}{4 x ^{2}}

q = - \frac{Q}{4}

The negative sign shows the central charge must be negative.

CBSE Class 12 Board · 2005 · All India · Subjective

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