Capacitance of parallel plate capacitor is …
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Capacitance of parallel plate capacitor is 50PF and distance between plates is 4mm it is charged to 200V and charging battery is removed Now a dielectric slab (k=4) of thickness 2mm is introduced between plates find the final charge and potential difference between plates also find energy loss
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Sia ? 4 years ago
Given:
To Find:
(i) final charge on each plate
(ii) finial potential difference between the plates
(iii) final energy is the capacitor
Solution:
Capacitance is given by
When dielectric slab constant is introduced
Capacitance is given by
putting the value of
Now,
(i) final charge on each plate
Thus, the final charge
(ii) finial potential difference between the plates
After charging the battery we removed the battery so the charge will remains same
(iii) final energy is the capacitor
Thus, Final energy of the capacitor
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