Electrostatics Question 214
Question: Equal charges $ q $ are placed at the four corners $ A,B,C,D $ of a square of length $ a $ . The magnitude of the force on the charge at B will be [MP PMT 1994; DPMT 2001]
Options:
A) $ \frac{3q^{2}}{4\pi {\varepsilon _{0}}a^{2}} $
B) $ \frac{4q^{2}}{4\pi {\varepsilon _{0}}a^{2}} $
C) $ ( \frac{1+2\sqrt{2}}{2} )\frac{q^{2}}{4\pi {\varepsilon _{0}}a^{2}} $
D) $ ( 2+\frac{1}{\sqrt{2}} )\frac{q^{2}}{4\pi {\varepsilon _{0}}a^{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
After following the guidelines mentioned above $ F _{net}=F _{AC}+F _{D}=\sqrt{F _{A}^{2}+F _{C}^{2}+2F _{A}F _{C}\cos\theta} $ Since $ F _{A}=F _{C}=\frac{kq^{2}}{a^{2}} $ and $ F _{D}=\frac{kq^{2}}{{{(a\sqrt{2})}^{2}}} $
$ F _{net}=\frac{\sqrt{2}kq^{2}}{a^{2}}+\frac{kq^{2}}{2a^{2}}=\frac{kq^{2}}{a^{2}}( \sqrt{2}+\frac{1}{2} ) $
$ =\frac{q^{2}}{4\pi {\varepsilon _{0}}a^{2}}( \frac{1+2\sqrt{2}}{2} ) $
BETA
