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Question: The electrostatic potential energy of two point charges, \(1\mu C\)each, placed 1 meter apart in air...

The electrostatic potential energy of two point charges, 1μC1\mu Ceach, placed 1 meter apart in air is?
A. 9×103J9\times {{10}^{3}}J
B. 9×109J9\times {{10}^{9}}J
C. 9×10−3J9\times {{10}^{-3}}J
D. 9×10−3eV9\times {{10}^{-3}}eV

Explanation

Solution

An electric potential which is also known as the electric field potential, potential decrease, or electrostatic potential is the net – work done to move a unit charge from a point to a particular point under the influence of some electric field with zero acceleration. This question can easily be solved by using a simple formula of electrostatic potential energy between the two point charges. This formula can easily be derived by using Coulomb's law.

Formula used:
For solving the given question, we will be using the formula of Electrostatic potential energy, i.e.,
Electrostatic potential energy =U=kq1q2r=U=\dfrac{k{{q}_{1}}{{q}_{2}}}{r}

Complete step-by-step answer:


By using the above-given formula for Electrostatic potential energy, we have
Electrostatic potential energy =U=kq1q2r=U=\dfrac{k{{q}_{1}}{{q}_{2}}}{r}
Where k is the Coulomb’s constant, k=8.9875517923×109 kg⋅m3⋅s−2⋅C−2k=8.9875517923\times {{10}^{9}}~kg\cdot {{m}^{3}}\cdot {{s}^{-}}^{2}\cdot {{C}^{-}}^{2}
For the sake of calculation, we will use k=9×109 kg⋅m3⋅s−2⋅C−2k=9\times {{10}^{9}}~kg\cdot {{m}^{3}}\cdot {{s}^{-}}^{2}\cdot {{C}^{-}}^{2}
And, q1{{q}_{1}} and q2{{q}_{2}} are the two point charges in this case,
q1=q2=1μC{{q}_{1}}={{q}_{2}}=1\mu C
Now,
q1=q2=1×10−6C{{q}_{1}}={{q}_{2}}=1\times {{10}^{-6}}C
And, r is the distance between the two point charges, in this case
r=1mr=1m
Now, by using the given data in the electrostatic potential energy formula
U=kq1q2rU=\dfrac{k{{q}_{1}}{{q}_{2}}}{r}
⇒U=9×109×1×10−6×1×10−61J\Rightarrow U=\dfrac{9\times {{10}^{9}}\times 1\times {{10}^{-6}}\times 1\times {{10}^{-6}}}{1}J
⇒U=9×109×1×10−121J\Rightarrow U=\dfrac{9\times {{10}^{9}}\times 1\times {{10}^{-12}}}{1}J
⇒U=9×109−12×11J\Rightarrow U=\dfrac{9\times {{10}^{9-12}}\times 1}{1}J
⇒U=9×10−3J\Rightarrow U=9\times {{10}^{-3}}J
So, the electrostatic potential energy of two point charges, 1μC1\mu C each, placed 1 meter apart in air is 9×10−3J9\times {{10}^{-3}}J

So, the correct answer is “Option C”.

Note: A coulomb is defined as the volume of electricity transported by a current of one ampere in a single second.
1C=6.24×1018electrons1C=6.24\times {{10}^{18}}electrons(Approximately)
The unit is named after Charles-Augustin de Coulomb, a French physicist who was active during the late 19th to mid 20th century. Electrostatic Potential energy is the reason why lightning happens.