Question

A charge Q1 = 1C and Q2 = -4C are fixed at a separation of 1 Metre. where should we place the third charge q so that the whole system is in equilibrium?

Answer 1

let distance be x
answer will be 2/3m

Answer 2

Answer:

The system is in equilibrium if the third charge q is place at a distance $\frac{1}{3} \: m$.

Explanation:

Given,Two charges are Q₁ = 1C and Q₂ = − 4C

and distance between them is 1 m.

Now we place the third charge q.

Let distance between Q₁ and q be x.

So distance between Q₂ and q is (1-x) m.

Now force between Q₁ and q $= k \frac{Q_1 \times q}{ {x}^{2} } = k \frac{1 \times q}{ {x}^{2} } = k \frac{q}{ {x}^{2} }$

Nand force between Q₂ and q $= k \frac{Q₂ \times q}{ {(1 - x)}^{2} } = k \frac{ - 4 \times q}{ {(1 - x)}^{2} }$N

The system is in equilibrium, if the total force acting on q is zero.

So,$k \frac{q}{ {x}^{2} } + k \frac{( - 4q)}{ {(1 - x)}^{2} } = 0$

$\frac{1}{ {x}^{2} } - \frac{4}{ {(1 - x)}^{2} } = 0$

$\frac{1}{ {x}^{2} } = \frac{4}{ {(1 - x)}^{2} } \\ \frac{1}{x} = \frac{2}{1 - x}$

By cross multiplication,

$1 - x = 2x$

$2x + x = 1$

$x = \frac{1}{3}$

The system is in equilibrium if the third charge q is place at a distance $\frac{1}{3} m$.

Here K =$\frac{1}{4\pi\epsilon }$.