Question

A point charge of $20MC$ is situated at a point O. A and B are points $0.05m$ and $0.15m$ away from this charge. Calculate amount of work done to move an electron from B to A.

Answer 1

Hint: The question can be approached by using the concept of Electric potential energy of a charge. First, we need to calculate the Electric potential energy at each point A and B respectively. Adding the electric potential energies at both the points will give us the total work done in moving the electron.
Formula used:
$U = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{r}$
$W = {U_A} + {U_B}$

Complete answer:
It is given that initially, the electron is at A, $0.05m$ away from the point charge of $20MC$. The point charge exerts an attractive force on the electron, so the electron has to do some work to move it against the electric field of a point charge. This is the electric potential energy of the electron, ${U_A}$. It is given by
${U_A} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r_A}}}$
Where,
${\varepsilon _0}$ is the permittivity of free space
${q_1}$ and ${q_2}$ are the charges
${r_A}$ is the distance between the charges
We have the charges, ${q_1} = 20MC = 20 \times {10^6}C$ and charge of an electron, ${q_2} = 1.6 \times {10^{ - 19}}C$. The distance between the charges as, ${r_A} = 0.5m$ and the proportionality constant $\dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$.
Substituting these values in the above formula, we get
$\eqalign{
  & {U_A} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r_A}}} \cr
  & \Rightarrow {U_A} = \left( {9 \times {{10}^9}} \right) \times \dfrac{{20 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{0.5}} \cr
  & \Rightarrow {U_A} = \left( {9 \times {{10}^9}} \right) \times \dfrac{{32 \times {{10}^{ - 13}}}}{{5 \times {{10}^{ - 1}}}} \cr
  & \Rightarrow {U_A} = 5.76 \times {10^{ - 2}} \cr
  & \Rightarrow {U_A} = 57.6mJ \cr} $
Similarly, when the electron is at B, the force exerted can be written as
${U_B} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r_B}}}$
Here, all quantities are the same except for ${r_B} = 0.15m$. The force ${F_B}$ will be
$\eqalign{
  & {U_B} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r_B}}} \cr
  & \Rightarrow {U_B} = \left( {9 \times {{10}^9}} \right) \times \dfrac{{20 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{1.5}} \cr
  & \Rightarrow {U_B} = \left( {9 \times {{10}^9}} \right) \times \dfrac{{32 \times {{10}^{ - 13}}}}{{15 \times {{10}^{ - 1}}}} \cr
  & \Rightarrow {U_B} = 2.88 \times {10^{ - 2}} \cr
  & \Rightarrow {U_B} = 28.8mJ \cr} $
Now, the work done in moving the electron from A to B is given by the sum of the electric potential energies. Mathematically written as,
$\eqalign{
  & W = {U_A} + {U_B} \cr
  & \Rightarrow W = 57.6mJ + 19.2mJ \cr
  & \Rightarrow W = 76.8mJ \cr} $
Therefore, the total work done in bringing the electron from point A to point B, under the influence of point charge is $76.8mJ$.

Note:
Don’t confuse electric potential energy for electric potential. Electric potential energy is the work done by the charge against an electric field. While Electric potential is the difference in electric potential energy per unit charge. And the suffix M before coulombs denotes the factor ${10^6}$, while m denotes the factor ${10^{ - 3}}$.