Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by\[{{Q}_{0}},{{V}_{0}},{{E}_{0}}\]and\[{{U}_{0}}\] respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by Q, V, E and U related to the previous one as: (1) \[Q>{{Q}_{0}}\] (2)\[V>{{V}_{0}}\] (3) \[E>{{E}_{0}}\] (4)\[U>{{V}_{0}}\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 and correct
D) 1, 2 and 4 are correct
Correct Answer: C
Solution :
When dielectric slab is introduced capacity gets increased\[\left( C=\frac{K{{\varepsilon }_{0}}A}{d} \right),\]while potential difference remains unchanged. \[\therefore \] \[V={{V}_{0}},C>{{C}_{0}}\] Again, \[Q=CV\] \[\therefore \] \[Q>{{Q}_{0}}\] and \[U=\frac{1}{2}C{{V}^{2}}\] \[\therefore \] \[U<{{U}_{0}}\] Also,\[E=\frac{V}{d}\]but E and d both are unchaged. Therefore, \[E={{E}_{0}}\]
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