1.

A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. What change, in any will take place in (i) charge on the plates (ii) electric field intensity between the plates (iii) the capacitance of the capacitor, (iv) potential difference between the plates and (v) the energy stored in the capacitor? Justify your answer in each case.​

Answer»

Given that,

A parallel plate capacitor is charged by a battery.

The battery is disconnected and a dielectric slab with its thickness equal to the plate SEPARATION is inserted between the plates.

(i). We NEED to calculate the charge

The battery is disconnected so the charge on capacitor remains constant.

(ii). We need to calculate the capacitance of the capacitor

The capacitance WITHOUT dielectric is

C=\dfrac{A\epsilon_{0}}{d}

When dielectric slab is inserted, then the capacitance

C'=\dfrac{kA\epsilon_{0}}{d}

C'=kC

Where, k = constant 

(iii) We need to calculate the ELECTRIC FIELD intensity between the plates

Using formula of electric field

E= \dfrac{V}{d}

Where, V = potential difference and

d = separation between the plates.

Potential difference decreases and d remains the same and electric field also decreases.

(iv). We need to calculate the potential difference

Using formula of potential difference

V=\dfrac{Q}{C}

As C increases and Q remains the same since the battery is disconnected, the potential difference between the plates decreases.

(v) We need to calculate the electric field intensity between the plates

Using formula of store energy

U=\dfrac{1}{2}\dfrac{Q^2}{C}

As Q is constant

So, Capacitance will be increases then stored energy will be decreases.

Hence, This is the required answer.


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