Question

Two point charges + 4q and +q are placed 30 cm

apart. At what point on the line joining them is the

electric field zero ? (from 4q) in cm​

Answer 1

EXPLANATION.

Two point charge = +4q and +q

Distance = 30 cm

To find where electric field is zero.

According to the question,

$\rm \to \: formula \: \: of \: \: coulombs \: \: law$

$\green{ \large{\bold{\rm \to \: \:f \: = \frac{k q_{1} q_{2} }{ {r}^{2} } }}}$

The electric field is zero at the point where the

forces exerted by charge +4q and +q on a

test charge Q are equal.

Therefore,

$\rm \to \: f_{1} \: = \: f_{2}$

1) = Force on +q point on a charge.

$\rm \to \: f_{1} \: = \frac{kqQ}{ x {}^{2} }$

2) = Force on +4q point on a charge.

$\rm \to \: f_{2} \: = \frac{4kqQ}{(l \: - \: x \: ) {}^{2} }$

=> Fnet = F¹ = F²

$\rm \: \to \: \frac{kqQ}{ {x}^{2} } = \frac{4kqQ}{(l \: - \: x \: ) {}^{2} }$

$\rm \to \: \frac{1}{x {}^{2} } = \frac{4}{(l \: - \: x \: ) {}^{2} }$

$\rm \to \: \frac{(l \: - \: x \: ) {}^{2} }{ {x}^{2} } = 4$

$\rm \to \: \frac{(l \: - \: x \: )}{x} = \sqrt{4}$

$\rm \to \: \frac{(l \: - \: x \: )}{x} = \pm \: 2$

Case = 1.

$\rm \to \: \frac{( \: l \: - \: x)}{x} = 2$

$\rm \to \: l \: - \: x \: = 2x$

$\rm \to \: l \: = 3x$

=> L = Length = 30 cm [ Given ]

put the value of L = 30 cm

$\rm \to \: 30 = \: 3x$

$\rm \to \: x \: = 10 \: cm$

Case = 2.

$\rm \to \: \frac{(l \: - \: x \: )}{x} = - 2$

$\rm \to \: l \: - x \: = - 2x$

$\rm \to \: l \: = - x$

$\rm \to \: x \: \ne \: - 10$

Therefore,

=> X = 10 cm.

=> ( L - x) = 30 - 10 = 20 cm

=> x = 10 cm