Question

Three particles, each having a charge of $10 \mu C$ are placed at the corners of an equilateral triangle of side $10 cm$. The electric potential energy of the system is:
(Given, $\dfrac{1}{{4\pi {\varepsilon _o}}} = 9 \times {10^9}{{ N}}{{.}}{{{m}}^2}/{C^2}$ )
A) $Zero$
B) $Infinite$
C) $27 J$
D) $100 J$

Answer 1

Hint: Given that three charged particles are placed at the corners of an equilateral triangle. The length of the sides of the triangle is given and the charges are known. Therefore we can find the electric potential energy due to any two charges at a time. Lastly, to find the total electric potential of the system, we have to find the sum of the electric potentials.

Formula used:
Potential energy due to a particle carrying charge Q is given by
$E = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{Q}{r}$, where $\dfrac{1}{{4\pi {\varepsilon _o}}} = 9 \times {10^9}{{ N}}{{.}}{{{m}}^2}/{C^2}$

Complete step by step solution:
Three particles are placed at the corners A, B and C of an equilateral triangle of side 10 cm. Each of the particles carries a charge of $10 \mu C$.
$Q = 10 \mu C$
Distance between any two charges, r = 10 cm = 0.1 m
Now, potential energy due to a particle carrying charge Q is given by
$\Rightarrow E = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{Q}{r}$, where $\dfrac{1}{{4\pi {\varepsilon _o}}} = 9 \times {10^9}{{ N}}{{.}}{{{m}}^2}/{C^2}$
Let, potential energy of the charges at A and B be given by ${E_{AB}}$
$\Rightarrow {E_{AB}} = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{{Q^2}}}{r}$
$ \Rightarrow 9 \times {10^9} \times \dfrac{{{{\left( {10 \times {{10}^{ - 6}}} \right)}^2}}}{{0.1}}$ J (substituting$\dfrac{1}{{4\pi {\varepsilon _o}}} = 9 \times {10^9}{{ N}}{{.}}{{{m}}^2}/{C^2}$)
$ \Rightarrow 9 \times {10^9} \times \dfrac{{{{10}^{ - 10}}}}{{0.1}}$ $J$
$ \Rightarrow 9 \times {10^9} \times {10^{ - 9}}$ $J$
$ \Rightarrow 9$ $J$
​Similarly, we can show that ${{{E}}_{{{BC}}}} = {{{E}}_{{{AC}}}} = {{9 J}}$
Therefore, total potential energy of the system is given by,
$\Rightarrow {E_{total}} = {E_{AB}} + {E_{BC}} + {E_{AC}} = \left( {9 + 9 + 9} \right){{ J}} = {{ }}27{{ J}}$

Hence, the correct answer is option (C).

Note: We know, potential energy due to a particle carrying charge Q is given by
$E = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{Q}{r}$, where $\dfrac{1}{{4\pi {\varepsilon _o}}} = 9 \times {10^9}{{ N}}{{.}}{{{m}}^2}/{C^2}$.
Note that in the given data, Coulomb is an SI unit while the unit of length is given in CGS. Therefore we will have to convert all the units either in CGS or in SI.