Question

Two-point charges A and B of values 15 pc and 10 pc are kept 20cm apart in air. Calculate the work done when the charge B is moved by 4cm towards charge A. When the distance between the charges is $r = 20cm$, the potential energy of the system is

Answer 1

Hint: To find the required solution of the given question deduce the expression of initial and final potential energy then find the work done by equating the expressions of initial and final potential energy.
Formula Used:
$U = \dfrac{1}{{4\pi {\varepsilon _ \circ }}} \times \dfrac{{{q_1}{q_2}}}{r}$

Complete answer:
An electric potential energy is defined as the charge placed in an electric field possesses potential energy and is measured by the work done in moving the charge from infinity to that point against the electric field.
Mathematically, it can be given as
$U = \dfrac{1}{{4\pi {\varepsilon _ \circ }}} \times \dfrac{{{q_1}{q_2}}}{r}$
Where ${q_1}$ and ${q_2}$ are the two charges separated by a distance ‘r’ and ${\varepsilon _ \circ }$ is the permittivity of free space or absolute permittivity or electric constant.
Initially, potential energy will be taken as,
${U_i} = \dfrac{1}{{4\pi {\varepsilon _ \circ }}} \times \dfrac{{{q_1}{q_2}}}{{{r_1}}}$
Final potential energy will be taken as,
${U_f} = \dfrac{1}{{4\pi {\varepsilon _ \circ }}} \times \dfrac{{{q_1}{q_2}}}{{{r_2}}}$
The change in potential energy $\left( {\Delta U} \right)$ is equivalent to work done if the given system does not have kinetic energy initially and finally. Work done is given as the difference between initial and final potential energy.
Mathematically, it can be given as
Work done = final P.E – Initial P.E
Here, ${r_1} = 20 \times {10^{ - 2}}m$ and since the charge B is moved by 4cm towards charge A.
So, ${r_2} = 16 \times {10^{ - 2}}m$
$w = \dfrac{1}{{4\pi {\varepsilon _ \circ }}} \times {q_1}{q_2}\left[ {\dfrac{1}{{16 \times {{10}^{ - 2}}}}} \right] - \dfrac{1}{{4\pi {\varepsilon _ \circ }}} \times {q_1}{q_2}\left[ {\dfrac{1}{{20 \times {{10}^{ - 2}}}}} \right]$
$ \Rightarrow w = \dfrac{1}{{4\pi {\varepsilon _ \circ }}} \times {q_1}{q_2}\left[ {\dfrac{1}{{16 \times {{10}^{ - 2}}}} - \dfrac{1}{{20 \times {{10}^{ - 2}}}}} \right]$
$ \Rightarrow w = \dfrac{{9 \times {{10}^9} \times 15 \times {{10}^{ - 12}} \times 10 \times {{10}^{ - 12}}}}{{{{10}^{ - 2}}}}\left[ {\dfrac{1}{{16}} - \dfrac{1}{{20}}} \right]$
$ \Rightarrow w = 135 \times 0.0125 \times {10^{ - 13}}$
$ \Rightarrow w = 1.6875 \times {10^{ - 13}}$joule
Hence, the potential energy of the system is $1.6875 \times {10^{ - 13}}$joule.

Note:
Potential energy and work done are the same thing as much as kinetic energy and work done are the same thing. Electric potential is defined as the energy which is needed to move a charge against an electric field. The electric potential is defined as the difference in the potential energy per unit charge between two points in an electric field.