JEE Main & Advanced
JEE Main Paper (Held On 10-Jan-2019 Evening)
question_answer
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is:
[JEE Main Online Paper (Held On 10-Jan-2019 Evening]
A)508 pJ
B)692 pJ
C)560 pJ
D) 600 pJ
Correct Answer:
A
Solution :
Internal energy = u \[\begin{align} & Q=CV\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{u}_{i}}=\frac{1}{2}\times 12\times {{10}^{-12}}\times 100 \\ & =12\times {{10}^{-12}}\,\times 10\,\,\,\,\,\,=600\,\times \,10-12\,\, \\ & =12\times {{10}^{-11}}\,J\,\,\,\,\,\,\,\,\,\,\,\,\,=6\times {{10}^{-10}}\,J \\ & \\ \end{align}\] After insertion \[C=KC=6.5\times 12\times {{10}^{-12}}\] Final energy \[{{u}_{f}}=\frac{{{Q}^{2}}}{2C'}\] \[\begin{align} & =\,\,\frac{12\times 12\times {{10}^{-11}}\,\times \,{{10}^{-11}}}{2\,\times \,6.5\,\times 12\times {{10}^{-12}}} \\ & \,\,\,\,\,\,\,\,\,\, \\ \end{align}\] So energy dissipated = \[=\,\,{{u}_{i}}\,=\,{{u}_{f}}\] \[\Rightarrow \text{ }508\text{ }pJ\]