InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers `1//3` of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is `C_1`. When the capacitor is charged, the plate area covered by the dielectric gets charge `Q_1` and the rest of the area gets charge `Q_2`. The electric field in the dielectric is `E_1` and that in the other portion is `E_2`. Choose the correct option/options, ignoring edge effects. A. `(E_(1))/(E_(2)) = 1`B. `(E_(1))/(E_(2)) = (1)/(K)`C. `(Q_(1))/(Q_(2)) = (3)/(K)`D. `(C_(1))/(C_(2)) = (3 + K)/(K)` |
|
Answer» Correct Answer - A When capacitor is charged, both parts of capacitors have common potential difference `V`. So `E_(1) = (V)/(d) = E_(2)` or `(E_(1))/(E_(2)) = 1` `C_(1) = *K (in_(0) A//3)/(d) , C_(2) = (in_(0) 2 A//3)/(d)` `C = C_(1) + C_(2) = (K in_(0) A)/(3d) + (2 in_(0) A)/(3d) + ((k-2) in_(0) A)/(3d)` `(C )/(C_(1)) = (k + 2)/(K)` `Q_(1) = C_(1) V = (K in_(0) A)/(3d)` `Q_(2) = C_(2)V = (2 in_(0) A)/(3d) V` `(Q_(1))/(Q_(2)) = (K)/(2)` |
|