Question

Two charges -2 C and 4 C are separated by 6 cm. Where should a third charge of 1 C be kept so that it is in equilibrium ?

Answer 1

$\mathtt{ \huge\fbox{Solution :)}}$

Given ,

• A and B be the point where -2 C and 4 C charge are located
• The distance b/w -2 C and 4 C = 6 cm

Let ,

• P be the point where 1 C charge is in equilibrium
• Distance b/w A and P = (x) cm
• Distance b/w P and B = (6 - x) cm

For the charge to remain in equilibrium it has to be palced in between -2C and 4C charge and the electrostatic force on it by the charge at A and B are equal and opposite

We know that , the electrostatic force between two charges is given by

$\large \mathtt{ \fbox{Force = k \frac{ q_{1} q_{2} }{ {(r)}^{2} } }}$

Thus ,

$\sf \hookrightarrow k \frac{(2 \times 1)}{ {(x)}^{2} } = k \frac{(1\times4) }{ {(6 - x)}^{2} } \\ \\ \sf \hookrightarrow \frac{2}{4} = \frac{ {(x)}^{2} }{ {(6 - x)}^{2} } \\ \\\sf \hookrightarrow {(6 - x)}^{2} = 2 {(x)}^{2} \\ \\ \sf \: \: \: squaring \: both \: sides \: \: we \: get \\ \\\sf \hookrightarrow 6 - x = \sqrt{2}x \\ \\ \sf \hookrightarrow 6 = \sqrt{2}x + x \\ \\\sf \hookrightarrow 6 = x( \sqrt{2} + 1) \\ \\\sf \hookrightarrow x = \frac{6}{ 2.414} \\ \\ \sf \hookrightarrow x = 2.4 \: \: cm$

Hence , for equibarium , the charge 1 C must be palced at a distance of 2.4 cm from the charge -2C

______________ Keep Smiling ☺

Answer 2

Explanation:

aise question me Fac=Fcb......