A) \[2.0\times {{10}^{-6}}J\]
B) \[2.4\times {{10}^{-5}}\]
C) \[0.8\times {{10}^{-5}}J\]
D) \[1.6\times {{10}^{-5}}J\]
Correct Answer: D
Solution :
The capacitance of a parallel plate capacitor of plate area A and plate separated by distance d is given by \[C=\frac{{{\varepsilon }_{0}}A}{d}\] Given, \[A=200\,c{{m}^{2}}=200\times {{10}^{-4}}{{m}^{2}}\], \[d=0.05\,cm=0.05\times {{10}^{-2}}m\], \[{{\varepsilon }_{0}}=8.86\times {{10}^{-12}}{{C}^{2}}/N-{{m}^{2}}\] \[\therefore \] \[C=\frac{8.86\times {{10}^{-12}}\times 200\times {{10}^{-4}}}{0.05\times {{10}^{-2}}}\] \[=3.54\times {{10}^{-10}}\mu F\] Energy stored in capacitor is \[U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times 3.54\times {{10}^{-10}}\times 9\times {{10}^{4}}\] \[=1.6\times {{10}^{-5}}J\] Note: This energy resides in the electric field created between the plates of the charged capacitor.
You need to login to perform this action.
You will be redirected in
3 sec
