Question

What is the force between two small charges of 2 x 10-7C placed 30 cm apart in air?
​

Answer 1

Given

:

Two small charges of  are placed 30 cm apart in the air.

To be found :

The force between two charges

• Theory:

According to Columb's law  :

$\tt F\propto Q_{1} \times Q_{2}...(1)$

And $\tt F\propto \frac{1}{r^2}....(2)$

Then,

$\bf F = \frac{kQ_{1}\ Q_{2}}{r^2}$

Solution :

We have

$\tt Q_{1}=2 \times 10^{-7}C$

$\tt Q_{2}=2 \times 10^{-7}C$

And

$\tt r = 0.3m$

By Columb's law

$\bf F= \frac{kQ_{1}\ Q_{2}}{r^2}$

Put the given values

$\sf F=\frac{k 2 \times 10^{-7} \times 2 \times10^{-7}}{3^2 \times 10^{-2}}$

$\sf F=\frac{k \times 4 \times 10^{-14} \times 10^2}{9}$

We know that

$\sf k=9 \times 10^9$

$\sf F=\frac{9 \times 10^9 \times 4 \times10^{-14} \times10^2}{9}$

$\sf F=4 \times 10^{-3}N$

Hence,

The force between two given charges is $\sf 4 \times 10^{-3}N$

More About the topic

1) It is a conservative force.

2) it depends upon medium.

3) It means inverse square law

4) It is a long range force.

Answer 2

Given ,

The two charges of same sign and magnitude 2 × 10^(-7) C are placed 30 cm apart in air

We know that , the force b/w two charges is given by

$\boxed{ \tt{F = k\frac{ q_{1}q_{2}}{ {(r)}^{2} } }}$

Thus ,

$\sf \implies F = \frac{9 \times {(10)}^{9} \times 2 \times {(10)}^{ - 7} \times 2 \times {(10)}^{ - 7} }{ { \{3 \times {(10)}^{ - 1} \} }^{2} }$

$\sf \implies F = \frac{9 \times 4 \times {(10)}^{ - 5} }{9 \times {(10)}^{ - 2} }$

$\sf \implies F = 4 \times {(10)}^{ - 3} \: \: N$

Therefore , the force between two given charges is 4 × 10^(-3) N