Electrostatics Question 383
Question: Two spheres of radius $ a $ and $ b $ respectively are charged and joined by a wire. The ratio of electric field of the spheres is [CPMT 1999; JIPMER 2000; RPET 2000]
Options:
A) $ a/b $
B) $ b/a $
C) $ a^{2}/b^{2} $
D) $ b^{2}/a $
Show Answer
Answer:
Correct Answer: B
Solution:
Joined by a wire means they are at the same potential.
For same potential $ \frac{kQ _{1}}{a _{1}}=\frac{kQ _{2}}{a _{2}} $
therefore $ \frac{Q _{1}}{Q _{2}}=\frac{a}{b} $
Further, the electric field at the surface of the sphere having radius R and charge Q is
$ \frac{E _{1}}{E _{2}}=\frac{kQ _{1}/a^{2}}{kQ _{2}/b _{2}}=\frac{Q _{1}}{Q _{2}}\times \frac{b^{2}}{a^{2}}=\frac{b}{a} $
BETA
