Direction (Q. Nos. 25) A parallel plate capacitor is as shown in figure. Half of the region in between the plates of the capacitor is filled with a dielectric material of dielectric constant K and in the remaining half air is present. The capacitor is given a charge Q with the help of a battery. Some surfaces are marked on the figure. |
Surface I is the right half inner surface of the upper plate of capacitor in tire region in which air is present. |
Surface II is the left half inner surface of the upper plate of capacitor. |
Surface III is the surface of the dielectric slab which is near to the upper plate of capacitor. |
Based on above information answer the following questions: |
A) \[\frac{KQ}{1+K}\]
B) \[\frac{{{K}^{2}}Q}{1+K}\]
C) \[\frac{-{{K}^{2}}Q}{1+K}\]
D) \[\frac{-K(K-1)Q}{1+K}\]
Correct Answer: D
Solution :
The charge on capacitor of capacitance \[K{{C}_{0}}\] is \[{{Q}_{1}}=\frac{QK{{C}_{0}}}{K{{C}_{0}}+{{C}_{0}}}=\left( \frac{QK}{1+K} \right)\]. This charge on the capacitor is shared by surfaces II and III. Let, charge on surface II be \[{{q}_{1}}\] and on III, it be \[{{q}_{2}}\]. \[{{q}_{1}}+{{q}_{2}}={{Q}_{1}}\] And from polarisation concept, \[{{q}_{1}}=-{{q}_{2}}\,\left( 1-\frac{1}{K} \right)\] Solving above equations, we get \[{{q}_{1}}=-\frac{-K(K-1)Q}{1+K}\]You need to login to perform this action.
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