Question

Or, There are two 6 micro F capacitors connected in parallel with a 6V d.c. source.
The first capacitor CA is kept connected while the second capacitor CB is
disconnected from the source. Now a mica sheet of dielectric constant 6 is inserted
inside to both the capacitors. Find the charges and electrostatic potential energies
in the two.​

Answer 1

Answer:

Capacitor which is connected to the battery

$Q = 2.16 \times 10^{-4} C$

$U = 6.48 \times 10^{-4} J$

capacitor which is disconnected from the battery

$Q = 3.6 \times 10^{-5} C$

$U = 1.8 \times 10^{-5} J$

Explanation:

When capacitor is connected across the battery then its potential remains constant and if it is not connected with the battery then its charge will remain the same

so here the capacitor which is still connected across the battery then it has

$Q = kCV$

$Q = 6(6 \times 10^{-6})(6)$

$Q = 2.16 \times 10^{-4} C$

Now energy stored in it is given as

$U = \frac{1}{2}(kC)V^2$

$U = \frac{1}[2}(6(6 \times 10^{-6}))(6^2)$

$U = 6.48 \times 10^{-4} J$

Now capacitor which is disconnected from the battery

$Q = CV$

$Q = (6 \times 10^{-6})(6)$

$Q = 3.6 \times 10^{-5} C$

Now energy of the capacitor

$U = \frac{Q^2}{2kC}$

$U = \frac{(3.6 \times 10^{-5})^2}{2(6 \times 6 \times 10^{-6}}$

$U = 1.8 \times 10^{-5} J$

#Learn

Topic : Capacitor

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