Electrostatics Question 246
Question: A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at $ x=+1 $ cm and C be the point on the y-axis at $ y=+1 $ cm. Then the potentials at the points A, B and C satisfy [IIT-JEE (Screening) 2001]
Options:
A) $ {{( \frac{q^{2}L}{2\pi {\varepsilon _{0}}mg} )}^{\frac{1}{3}}} $
B) $ {{( \frac{qL^{2}}{2\pi {\varepsilon _{0}}mg} )}^{\frac{1}{3}}} $
C) $ {{( \frac{q^{2}L^{2}}{4\pi {\varepsilon _{0}}mg} )}^{\frac{1}{3}}} $
D) $ {{( \frac{q^{2}L}{4\pi {\varepsilon _{0}}mg} )}^{\frac{1}{3}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
In equilibrium $ F _{e}=T\text{sin}\theta $ ……. (i) $ mg=T\text{cos}\theta $ ……. (ii) $ \tan \theta =\frac{F _{e}}{mg}=\frac{q^{2}}{4\pi {\varepsilon _{o}}x^{2}\times mg} $ also $ \tan \theta \approx \sin \theta =\frac{x/2}{L} $ Hence $ \frac{x}{2L}=\frac{q^{2}}{4\pi {\varepsilon _{o}}x^{2}\times mg} $
therefore $ x^{3}=\frac{2q^{2}L}{4\pi {\varepsilon _{o}}mg} $
therefore $ x={{( \frac{q^{2}L}{2\pi {\varepsilon _{o}}mg} )}^{1/3}} $
