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2. Electric Potential and Capacitance
hard
The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant $K $ is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.
A
the energy stored in the capacitor will become $K$ -times
B
the force of attraction between the plates will increase to $K^2$-times
C
the charge on the capacitor will increase to $K$-times
D
all of the above
Solution
$U_{i}=\frac{1}{2} C V^{2} \& U_{f}=\frac{1}{2}(K C) V^{2}$
$U_{f}=K U_{i}$
Since charge on each plate increase to $k$ times so force increase to $k^{2}$ times
$Q_{i}=C V \& Q_{f}=(K C) V$
Standard 12
Physics
