A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them. When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitudes of the capacitance and the potential difference between the plates are

Answer :

The capacity increases and potential decreases with the introduction of dielectric .

Explanation:

The capacity of parallel plate capacitor when air is in between

C₀ = ε₀A/d

and when dielectric is introduced

The capacity C = Kε₀A/d

where K is dielectric constant .

When dielectric is introduced between the parallel plate capacitor , its capacity increases .

If the capacity of capacitor is C₀ initially . It will be 3 C₀ , after introducing the dielectric .

Thus if the initial potential V₀ = [tex]\frac{Q_0}{C_0}[/tex] , the potential with dielectric is = [tex]\frac{Q_0}{3C_0}[/tex]

Thus the potential will become become 1/3 of initial value .

and the capacitance becomes 3 times the initial value .

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