Question

Two charges of magnitude $5nC$ and $- 2nC$, one placed at point (2cm, 0, 0) and (x cm, 0, 0) in a region of space, where there is no other external field. If the electrostatic potential energy of the system is $- 0.5\mu J$. The value of x is:
A. 20 cm
B. 80 Cm
C. 4 cm
D. 16cm

Answer 1

In question, the numerical values of magnitude of charges and electrostatic potential energy of the system is given, firstly, by distance formula, ‘r’ is derived in terms of x. Thus by substituting values in the mathematical formula of potential energy, numerical value of ‘x’ can be calculated.

Formula Used:

1. Distance between two points = $\sqrt {{{({x_B} - {x_A})}^2} + {{({y_B} - {y_A})}^2} + {{({z_B} - {z_A})}^2}}$
2. $U = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{{{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}}}}{r}$
   Where, $U$= electrostatic potential energy
   $\dfrac{1}{{4\pi {\varepsilon _0}}}$= constant of proportionality (value is$9 \times {10^9}N{m^2}/{C^2}$)
   ${\varepsilon _0}$= absolute electric permittivity of free space or vacuum
   $r$ = distance between the points where electric charges are placed.

Complete step by step answer:
Given that there are two charges placed in a region of space with magnitude $5nC$and$- 2nC$.Let us assume that ‘${{\text{q}}_{\text{1}}}$’ be the first charge whose magnitude is$5nC$and placed at (2cm, 0, 0).
${{\text{q}}_{\text{1}}} = 5nC$= $5 \times {10^{ - 9}}C$
And ‘${{\text{q}}_{\text{2}}}$’ be another charged particle whose magnitude is $- 2nC$ and placed at (x cm, 0, 0).
${{\text{q}}_{\text{2}}} = - 2nC$= $- 2 \times {10^{ - 9}}C$
Also, the electrostatic potential energy of the system is given which is equal to$- 0.5\mu J$.
$U = - 0.5 \times {10^{ - 6}}$ J
In electrostatics, electrostatic potential energy is defined as the amount of energy possessed by the virtue of two charges placed at some fixed points and mathematically it can be calculated by:
$U = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{{{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}}}}{r}$…………………………………Eq.1
In space, the position of charge ‘${{\text{q}}_{\text{1}}}$’ is given as (2cm, 0, 0) and position of charge ‘${{\text{q}}_{\text{2}}}$’ as (x cm, 0, 0). So these are two points and by distance formula we would find the value of ‘r’ in terms of ‘x’ and substitute in equation 1 to find the value of X. Thus, applying, distance formula,
$\Rightarrow$ Distance between two points = $\sqrt {{{({x_B} - {x_A})}^2} + {{({y_B} - {y_A})}^2} + {{({z_B} - {z_A})}^2}}$

\sqrt {{{({x_B} - {x_A})}^2} + {{({y_B} - {y_A})}^2} + {{({z_B} - {z_A})}^2}} \\\ \Rightarrow \sqrt {{{(x - 2)}^2} + {{(0 - 0)}^2} + {{(0 - 0)}^2}} \\\ \Rightarrow \sqrt {{{(x - 2)}^2}} \\\ \Rightarrow x - 2 \\\ $$ Now, substitute all the values in equation 1, we get $$ \Rightarrow - 0.5 \times {10^{ - 6}} = 9 \times {10^9} \times \dfrac{{5 \times {{10}^{ - 9}} \times - 2 \times {{10}^{ - 9}}}}{{(x - 2) \times {{10}^{ - 2}}}}$$

\Rightarrow (x - 2) = 9 \times {10^9} \times \dfrac{{5 \times {{10}^{ - 9}} \times - 2 \times {{10}^{ - 9}}}}{{{{10}^{ - 2}} \times - 0.5 \times {{10}^{ - 6}}}} \\
\Rightarrow (x - 2) = \dfrac{{9 \times {{10}^{9 - 9 - 9 + 2 + 6 + 2}}}}{5} \\
\Rightarrow x - 2 = 18 \\
\Rightarrow x = 18 + 2 \\
\therefore x = 20 $Therefore, the position coordinate of charge ‘${{\text{q}}_{\text{2}}}$$’ would be (20 cm, 0, 0).

Hence, the correct option is A.

Note: All the values should be converted in SI units to avoid any error.

$$1nC = {10^{ - 9}}C \\\ \Rightarrow 1\mu J = {10^{ - 6}}J \\\$$

Since, electrostatic potential is a scalar quantity, it would not have any direction just like an electric field which is a vector quantity. Also, this potential energy is inversely proportional to the distance between the charges, therefore an increase in distance would decrease the energy of the system.