Electrostatics Question 794
Question: As per the diagram, a point charge +q is placed at the origin O. Work done in taking another point charge-Q from the point A [coordinates (0, a)] to another point B [coordinates (a, 0)] along the straight path AB is:
Options:
A) zero
B) $ ( \frac{-qQ}{4\pi {\varepsilon _{0}}}\frac{1}{a^{2}} )\sqrt{2}a $
C) $ ( \frac{-qQ}{4\pi {\varepsilon _{0}}}\frac{1}{a^{2}} ).\frac{a}{\sqrt{2}} $
D) $ ( \frac{qQ}{4\pi {\varepsilon _{0}}}\frac{1}{a^{2}} ).\sqrt{2}a $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We know that potential energy of two charge system is given by $ U=\frac{1}{4\pi {\in _{0}}}\frac{q _{1}q _{2}}{r} $ According to question, $ U _{A}=\frac{1}{4\pi {\varepsilon _{0}}}\frac{( +q )( -Q )}{a}=-\frac{1}{4\pi {\varepsilon _{0}}}\frac{Qq}{a} $
$ \text{and }U _{B}=\frac{1}{4\pi {\varepsilon _{0}}}\frac{( +q )( -Q )}{a}=-\frac{1}{4\pi {\varepsilon _{0}}}\frac{Qq}{a} $
$ \Delta U=U _{B}-U _{A}=0 $ We know that for conservative force, $ W=-\Delta U=0 $
BETA
