Name: Permittivity of Air and Free Space
Uploaded: 2019-07-19
Duration: 6 min 47 s
Description: What is meant by relative permittivity? In this video, you will get to know more about it. Watch this video and know the value and dimensional formula of the...

Question

Particle $(1)$ of charge $+ q$ and particle $(2)$ of charge $+ 9.00 q$ are held at separation, $L = 8.00 cm$ on an $x -$ axis. If particle $(3)$ of charge $q_{3}$ is to be located such that the three particles remain in place when released, then what must be the

$(A)$ $x$ co-ordinate of particle $(3)$

$(B)$ $y$ co-ordinate of particle $(3)$

$(C)$ The ratio $\frac{q_{3}}{q}$ ?

Accepted Answer

$+ q$ Let $q_{3} > 0$ , that means it is a positive charge.

If we put charge $q$ on $L = 8.00 cm$ , then it will experience a net force towards left.

If we place charge $q_{3}$ on $C$ , then it will experience a net force rightwards due to $+ q$ and $+ 9 q$ .

If we place $q_{3}$ at $B$ (anywhere between $+ q$ and $+ 9 q$ ), then $(3)$ and $(C)$ will experience a net force, so the system will not be in equilibrium.

Hence, we can't take $q_{3} > 0$ .

Let $q_{3} < 0$ , that means it is a negative charge $(- q_{3})$ .

Let us place $- q_{3}$ on $A$ , then $- q_{3}$ will experience a net charge.

If we place $- q_{3}$ on $C$ , then it experiences a net force.

Now, we place $- q_{3}$ in between $+ q$ and $+ 9 q$ .

$F_{+ q} =$ Force on $- q_{3}$ due to $+ q$ charge,

$F_{+ 9 q} =$ Force on $x -$ due to $+ 9 q$ charge.

From the above figure, net force becomes zero on $(A)$ .

Hence,

$F_{+ q} = F_{+ 9 q}$

$\Rightarrow \frac{k q q_{3}}{x^{2}} = \frac{k (9 q) q_{3}}{{(8 - x)}^{2}}$

$\Rightarrow \frac{1}{x} = \frac{3}{(8 - x)}$

$\Rightarrow 3 x = (8 - x)$

$\Rightarrow x = 2 cm$ .

$(B)$ No force by $+ q$ or $+ 9 q$ on $q_{3}$ , hence, the charge is applied along $y -$ axis.

$(C)$ In the above figure,

$F_{2} =$ Force on $+ 9 q$ due to $- q_{3}$ ,

$F_{1} =$ Force on $+ 9 q$ due to $+ q$ .

For an equilibrium of the system, the net force on $+ 9 q$ , must be zero.

Hence,

$F_{1} = F_{2}$

$\frac{k q (9 q)}{{(8)}^{2}} = \frac{k q_{3} (9 q)}{{(8 - x)}^{2}}$

$\Rightarrow \frac{\sqrt{q_{3}}}{(8 - x)} = \frac{\sqrt{q}}{(8)}$

$\Rightarrow 8 \sqrt{q_{3}} = 8 \sqrt{q} - (\sqrt{q}) x$ .

Substituting $x = 2 cm$ in the above equation, we get,

$8 \sqrt{q_{3}} = 8 \sqrt{q} - (\sqrt{q}) (2)$

$\Rightarrow 8 \sqrt{q_{3}} = 6 \sqrt{q}$

$\Rightarrow \frac{q_{3}}{q} = \frac{36}{64} = 0.563$ .

Important Questions on Coulomb's Law

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