Electrostatics Question 374
Question: Two insulated charged conducting spheres of radii $ 20cm $ and $ 15cm $ respectively and having an equal charge of $ 10C $ are connected by a copper wire and then they are separated. Then [MP PET 1997]
Options:
A) Both the spheres will have the same charge of $ 10C $
B) Surface charge density on the $ 20cm $ sphere will be greater than that on the $ 15cm $ sphere
C) Surface charge density on the $ 15cm $ sphere will be greater than that on the $ 20cm $ sphere
D) Surface charge density on the two spheres will be equal
Show Answer
Answer:
Correct Answer: C
Solution:
After redistribution, charges on them will be different, but they will acquire common potential i.e. $ k\frac{Q _{1}}{r _{1}}=k\frac{Q _{2}}{r _{2}} $
therefore $ \frac{Q _{1}}{Q _{2}}=\frac{r _{1}}{r _{2}} $ As $ \sigma =\frac{Q}{4\pi r^{2}} $
therefore $ \frac{{\sigma _{1}}}{{\sigma _{2}}}=\frac{Q _{1}}{Q _{2}}\times \frac{r _{2}^{2}}{r _{1}^{2}} $
therefore $ \frac{{\sigma _{1}}}{{\sigma _{2}}}=\frac{r _{2}}{r _{1}} $
therefore $ \sigma \propto \frac{1}{r} $ i.e. surface charge density on smaller sphere will be more.
BETA
