Electrostatics Question 281
Question: Each plate of capacitance C has partial value of charge [MP PMT 2003]
Options:
A) $ \frac{Q _{1}+Q _{2}+Q _{3}+Q _{4}}{2C} $
B) $ \frac{Q _{2}+Q _{3}}{2C} $
C) $ \frac{Q _{2}-Q _{3}}{2C} $
D) $ \frac{Q _{1}+Q _{4}}{2C} $
Show Answer
Answer:
Correct Answer: C
Solution:
Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal, $ Q _{2}=-Q _{3} $ . The charge on a capacitor means the charge on the inner surface of the positive plate (here it is $ Q _{2} $ ) Potential difference between the plates $ =\frac{charge}{capacitance} $
$ =\frac{Q _{2}}{C}=\frac{2Q _{2}}{2C} $
