Electrostatics Question 271
Question: given below shows two identical parallel plate capacitors connected to a battery with switch $ S $ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant 3. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric [IIT 1983]
Options:
A) 40
B) 32
C) 8
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
Suppose $ C=\text{ 8}\mu F,C’=\text{ 16}\mu F $ and V = 250 V, V’ = 1000V
Suppose m rows of given capacitors are connected in parallel and each row contains n capacitors then potential difference across each capacitor $ V=\frac{V’}{n} $
and equivalent capacitance of network $ C’=\frac{mC}{n} $ on putting the values we get n = 4 and m = 8 \ Total capacitors = nx m = 4x 8 = 32
Short Trick : For such type of problems number of capacitors = $ \frac{C’}{C}\times {{( \frac{V’}{V} )}^{2}}=\frac{16}{8}{{( \frac{1000}{250} )}^{2}}=32 $
