Question

14. An infinite number of charges, each of charge
placed on the x-axis with co-ordinates r = 1,2,4,8
If a charge of 1 C is kept at the origin, then what is the
force acting on 1 C charge?
​

Answer 1

Question :-

An infinite number of charges, each of charge $\sf 1 \mu \: C$ placed on the x-axis with co-ordinates r = 1,2,4,8 ....

If a charge of 1 C is kept at the origin, then what is the force acting on 1 C charge?

Answer :-

$\to \boxed{\sf f = 12 \times {10}^{3} } \: N \: \\ \sf or \\ \: \to \: \boxed{\sf \to f = 12000 \: N} \:$

To Find :-

→ Force acting on 1 C charge .

Explanation :-

Given that :

• Each of charge (q) = $\sf 1 \mu \: C$

• distance between them → 1 ,2 ,4 , ......

According to the question,

→ All charges are at a increasing distance of

→ 1 ,2 ,4 ,8 ..,..( an increasing GP )

We know that -

$\sf \mapsto f = \frac{1}{ 4 \pi \: ∈_0 } \frac{q_1 q_2}{ {r_{12}}^{2} }$

All charges are same ,

hence ,total Force acting on charges is -

$\mapsto \sf f = \frac{1 }{ 4 \pi \: ∈_0 \: } \big\{ \frac{1 \mu}{ {1}^{2} } + \frac{1 \mu}{ {2}^{2} } + \frac{1 \mu}{ {4}^{2} } + .... \infty \big \} \\ \\ \mapsto \sf f = \frac{1 \times {10}^{ - 6} }{ 4 \pi \: ∈_0 } \big \{ \frac{1}{1} + \frac{1}{ {2}^{2} } + \frac{1}{ {2}^{4} } + ... \infty \big \} \:$

We know that ,

Sum of infinite GP

$\to \boxed{ \sf \: s = \frac{a}{a - r} } \\ \\ \sf in \: this \: sequence \: \\ \: \\ \to \: \frac{1}{1} + \frac{1}{ {2}^{2} } + \frac{1}{ {2}^{4} } + .... \infty \: \\ \: \\ \: \: \: \: \: \: \: \: \: \implies \sf first \: term \: (a) = 1 \\ \\ \: \: \: \: \: \: \: \: \sf \implies \: common \: ratio \: (r) = \frac{1}{4} \\ \\ \therefore \\ \\ \to \sf \: f = \frac{1 \times {10}^{ - 6} }{ 4 \pi \: ∈_0 \: } \bigg( \frac{1}{ 1 - \frac{1}{4} } \bigg) \\ \\ \to \sf f = \frac{1 \times {10}^{ - 6} }{ 4 \pi \: ∈_0 \: } \bigg( \frac{1 }{ \frac{4 - 1}{4} } \bigg) \\ \\ \to \: \sf \: f = \frac{1 \times {10}^{ - 6} }{ 4 \pi \: ∈_0 \: } \bigg( \frac{4}{ 3} \bigg) \\ \\ \because \boxed{ \sf \frac{1}{ 4 \pi \: ∈_0 \: } = 9 \times {10}^{9} } \\ \\ \to \sf \: f = \frac{9 \times {10}^{9} \times 4 \times {10}^{ - 6} }{3} \\ \\ \to \boxed{\sf \: f = 12 \times {10}^{3} } \: N \: \\ \sf or \\ \: \to \: \boxed{\sf \: \to \: f = 12000 \: N}$