Question

Two point charges of +1.0 my C are kept stationary 2m apart.How much work is needed to be done to bring them 1m apart ?
1) 4.5 mJ
2) 9 mJ
3) 45 mJ
4) 90 mJ

Answer 1

4.5 mJ is the right answer

Answer 2

Answer:

The work done is 4.5 mJ.

1 is correct option

Explanation:

Given that,

Given that,

Charges $q =1.0\mu C=1\times10^{-6}\ C$

Distance d = 2 m

The potential energy is,

$P=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q^2}{r}$

The work done is equal to the change in potential energy.

$W=P_{f}-P_{i}$

Where, $P_{i}$=Initial potential energy

$P_{f}$=final potential energy

The initial potential energy is

$P_{i}=-\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q^2}{2}$

The final potential energy is

$P_{f}=-\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q^2}{1}$

Now,work done = change in potential energy

$W=P_{f}-P_{i}$

$W=-\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q^2}{1}-(-\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q^2}{2})$

$W=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q^2}{2}$

Put the value into the formula

$W=\dfrac{9\times10^{9}\times1.0\times10^{-6}}{2}$

$W=4.5\times10^{-3}\ J$

$W=4.5\ mJ$

Hence, The work done is 4.5 mJ.