Question

If two charges 5 using and -6 UC are placed at a distance of 20 CM then calculate the electric force working between them

Answer 1

The electric force acting between the charges is 135 N

Explanation:

Given:

Charges of 5 μC and -6 μC are placed at a distance of 20 cm

To find out:

The force between the charges

Solution:

Since the charges are of opposite sign, the force will be attractive

We know that the force between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by

$F=k\frac{q_1q_2}{r^2}$     where $k=9\times 10^9$ kg m³/s²/C²

Here

$q_1=5\times 10^{-6}$ C

$q_2=6\times 10^{-6}$ C

$r=20\times 10^{-2}$ m

Therefore, the force

$F=9\times 10^9\frac{5\times 10^{-6}\times 6\times 10^{-6}}{(20\times 10^{-2})^2}$

$\implies F=\frac{270\times 10^{-3}}{20\times 10^{-4}}$

$\implies F=135$ N

Hope this answer is helpful.

Know More:

Q: If two identical spheres having charge 16 uC and -8 uC are kept at certain distance apart,then the force is F. They are touched and again kept at the same distance then what will be d force ?

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