Question

An air capacitor is first charged through a battery. The charging battery is then removed and a electric slab of dielectric constant K=4 is inserted between the plates. Simultaneously, the distance between the plates is reduced to half, then find change, C, E, V and U.

Answer 1

Given :

An air capacitor is first charged through a battery

The dielectric constant = k = 4

The new distance between plates = d' = $\dfrac{d}{2}$

To Find :

The charge = Q  c

The electric field = E

The potential difference

The stored potential energy

Solution :

For of a capacitor = Capacitance  = $\dfrac{kA\varepsilon _o}{d}$

i.e                  C ∝ $\dfrac{k}{d}$

For,  change in capacitance = C' = $\dfrac{k}{d'}$

Or,                                             C' = $\dfrac{k}{\dfrac{d}{2} }$ = 8 C

∴   C' = 8 C

So, The change in  capacitance = C' = 8 C

Again

Change in electric field = E' = $\dfrac{E_o}{k}$

i.e                                       E' ∝ $\dfrac{1}{k}$

So,                                      E' = $\dfrac{1}{4}$ E

So, Change in electric field = E' = one - four times of initial value

Change in potential

 V = $\dfrac{q}{C}$

So,  V' = $\dfrac{q}{C'}$

Or,  V' = $\dfrac{q}{8C}$

So,  V' = $\dfrac{V}{8}$

So, Change is potential is one eight of initial potential

Change in potential energy

U' = $\dfrac{1}{2}$ $\dfrac{q^{2} }{C}$

i.e   U' ∝ $\dfrac{1}{C}$

∴   U' = $\dfrac{U}{8}$

So, Change in potential energy = $\dfrac{1}{8}$  times of initial value

Hence, The change in  capacitance is 8 times initial

 Change in electric field is one- fourth  of initial value

 Change is potential is one eight of initial potential  

 Change in potential energy one-eight of initial    Answer