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Question

Physics Question on electrostatic potential and capacitance

A capacitor having capacity of 2.0 μ2.0 \,\mu F is charged to 200 V200 \,V and then the plates of the capacitor are connected to a resistance wire. The heat produced in joule will be

A

2×10−22\times {{10}^{-2}}

B

4×10−24\times {{10}^{-2}}

C

4×1044\times {{10}^{4}}

D

4×10104\times {{10}^{10}}

Answer

4×10−24\times {{10}^{-2}}

Explanation

Solution

Heat produced in a wire is equal to energy stored in capacitor. H=12CV2H=\frac{1}{2}C{{V}^{2}} =12×(2×10−6)×(200)2=\frac{1}{2}\times (2\times {{10}^{-6}})\times {{(200)}^{2}} =10−6×200×200={{10}^{-6}}\times 200\times 200 =4×10−2J=4\times {{10}^{-2}}J