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Two identical capacitors of plate dimensions l xx b and plate separation 'd' have dielectric slabs filled in between the space of the plates as shown in Fig Obtain the relation between the dielectric constants K , K_1 and K_2. |
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Answer» Solution :For capacitor shown in FIG area of each plate of capacitor `A_1 = lb` and capacitance `C_1 = (K in_(0) A_(1))/(d)` The capacitor shown in Fig is equivalent to two capacitors each of plate area `A_(2) = (l)/(2).b = (lb)/(2) = (A_(1))/(2)` but FILLED with dielectrics of dielectric constants `K_1` and `K_2` respectively JOINED in series . Therefore , net capacitance of this arrangement is `C_(2) = (K_(1) in_(0) (A_(1)//2))/(d) + (K_(2) in_(0) (A_(1)//2))/(d) = (in_0 A_(1))/(d) [ (K_(1) + K_(2))/(2)]` Since `C_(1) = C_(2)` hence we CONCLUDE that `(K in_(0) A_(1))/(d) = (in_(0) A_(1))/(d) [ (K_(1) + K_(2))/(2)] implies K= (K_(1) + K_(2))/(2)` |
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