Question

Two point charges +16 microcoulomb and -9 microcoulomb are placed at 8 cm apart in air.you are asked to place a +10 microcoulomb at a third position such that the net force on +10 microcoulomb is zero .where will you place the charge

Answer 1

Given:

Two point charges +16 microcoulomb and -9 microcoulomb are placed at 8 cm apart in air.

A third charge of +10 micro-C is placed.

To find:

Position of the charge where it will experience zero force.

Calculation:

The charge has to placed closer to the -9 micro-C charge so as to gain equilibrium.

Distance of 10 micro-C from -9 micro-C be x.

$\therefore \: force1 = force2$

$= > \: \dfrac{k \times 16 \times {10}^{ - 6} \times 10 \times {10}^{ - 6} }{ {(8 + x) }^{2} } = \dfrac{k \times 9 \times {10}^{ - 6} \times 10 \times {10}^{ - 6} }{ {x }^{2} }$

Cancelling the common terms:

$= > \: \dfrac{ 16 }{ {(8 + x) }^{2} } = \dfrac{9 }{ {x }^{2} }$

Taking square root on both sides:

$= > \: \dfrac{ 4 }{ 8 + x } = \dfrac{3 }{ x }$

$= > \: 4x = 24 + 3x$

$= > \: x = 24 \: cm$

So, the 10 micro-C charge has to be placed 24 cm from the -9 micro-C charge and (24+8) = 32cm from +16 micro-C charge.