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Derive a relation for capacitance of a parallel plate capacitor with dielectric slab of thickness t filled between the plates. Or. Write expression for capacitance of a parallel plate capacitor and explain the effect of dielectrics on capacitance. |
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Answer» Solution :Capacitance of a parallel PLATE capacitor with vacuum between the two plates is `C_(0)=(epsi_(0)A)/(d)` Let a plate A be given +Q charge and plate B be given -Q charge. Introduce dielectric slab, say of thickness t(t `therefore`net electric field inside the dielectric is `vecE=vecE_(0)-vecE_(p)`. Potential difference between A and B is  But `(E_0)/(E)=K`, the dielectric constant. `thereforeE=(E_(0))/(K)` or `V=E_(0)(d-t)+(E_(0))/(K)xxt` or `V=E_(0)[d-t+(t)/(K)]` But `E_(0)=(sigma)/(epsi_(0))=(Q)/(Aepsi_(0))` `therefore V=(Q)/(Aepsi_(0))[d-t+(t)/(K)]` But capacity, `C=(Q)/(V)` `therefore C=(Q)/((Q)/(Aepsi_(0))[d-t+(t)/(K)])` or `Q=(Aepsi_(0))/(d-t+(t)/(K))` If whole of the space between plates is filled with dielectric, then t=d `therefore C=(Aepsi_(0))/(d-d+(d)/(K))` or `C=(KAepsi_(0))/(d)`. |
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