Question

Two charges 4 x 10-9C and -16 x 10-9 C are separated by a distance 20 cm in air. The position of the neutral point from the small charge is
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Answer 1

Given:

Two charges 4 x 10-9C and -16 x 10-9 C are separated by a distance 20 cm in air.

To find:

Position of neutral point from smaller point?

Calculation:

Neutral Point is at Point "N". Refer to diagram.

$\rm \: E1 = E2$

$\rm \implies\: \dfrac{k(4 \times {10}^{ - 9} )}{ {x}^{2} } = \dfrac{k(16 \times {10}^{ - 9}) }{ {(20 + x)}^{2} }$

$\rm \implies\: \dfrac{4}{ {x}^{2} } = \dfrac{16}{ {(20 + x)}^{2} }$

Taking Square Root:

$\rm \implies\: \dfrac{2}{x} = \dfrac{4}{20 + x}$

$\rm \implies\: 40 + 2x = 4x$

$\rm \implies\: 2x = 40$

$\rm \implies\: x = 20 \: cm$

So, neutral point is 20 cm from smaller charge.