Question

Two point charges 16 micro coulomb and - 9 mivro coulomb are placed 8cm apart in air. You are asked to place a 10 micro coulomb charge at third place such that force on this will be zero

Answer 1

Given :

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

To find :

Determine the Position of 10μC charge such that force on this will be zero .

Theory :

•Coulombs law

$\bf\:F =\dfrac{kq_{1}q_{2} }{r {}^{2} }$

Solution :

Let the point A has the positive charge +16 μC and the point B has the negative charge -9 μC.

Let the position of 10μC be P , x cm distance from point B .

P is the point where the electric force is zero.

The electric force acting on point P due to the positive charge at A is,

$\bf\:F_{PA}=\sf\:\frac{k\times10\times16\times10{}^{-6}\times10{}^{-6}}{(x+8){}^{2}\times10{}^{-4}}$

$= \frac{k \times 10 {}^{ - 12} \times 10 \times 16 }{(x + 8) {}^{2} \times 10 {}^{ - 4} }$

The electric force acting on point P due to the negative charge at B is,

$\bf\:F_{PB}=\sf\:\frac{k\times10\times9\times10{}^{-6}\times10{}^{-6}}{(8){}^{2}\times10{}^{-4}}$

$= \frac{k \times 10 {}^{ - 12} \times 9 \times 10 }{8 {}^{2} \times 10 {}^{ - 4} }$

Since force on 10μC at point P is zero

$\implies\:\sf\:F_{PA}=F_{PB}$

$\dfrac{k \times10 {}^{ - 12} \times10 \times 16 }{(x + 8) {}^{2} \times 10 { }^{ - 4} } = \dfrac{k \times 10 {}^{ - 12} \times 10 \times 9 }{8 {}^{2} \times 10 {}^{ - 4} }$

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

Taking square root on both sides

$\implies \frac{4}{x + 8} = \frac{3}{x}$

$\implies3(x + 8) = 4x$

$\implies \bf x = 24cm$

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Therefore,the Position of 10μC charge such that force on this will be zero is 24cm from -9μC charge .