Electrostatics Question 689
Question: Two pith balls carrying equal charges are suspended from a common point by strings of equal the balls now become
Options:
A) $ ( \frac{r}{\sqrt[3]{2}} ) $
B) $ ( \frac{2r}{\sqrt{3}} ) $
C) $ ( \frac{2r}{3} ) $
D) $ {{( \frac{r}{\sqrt{2}} )}^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \tan \theta =\frac{F _{e}}{mg}\Rightarrow \frac{r/2}{y}=\frac{\frac{kq^{2}}{r^{2}}}{mg} $
[$ \because F=\frac{kq^{2}}{r^{2}} $
from coulomb’s law] $ \Rightarrow r^{3}\propto y\Rightarrow r{{’}^{3}}\propto \frac{y}{2}\Rightarrow \frac{r’}{r}=\frac{1}{{{2}^{1/3}}}\Rightarrow r’=\frac{r}{\sqrt[3]{2}} $
BETA
