A) \[150V\]
B) \[100V\]
C) \[250V\]
D) \[300V\]
Correct Answer: C
Solution :
[c] \[A=200c{{m}^{2}}\] |
\[d=1.5cm\] |
\[F=25\times {{10}^{-6}}N\] |
\[\because E=\frac{\sigma }{2{{\in }_{o}}}=\frac{Q}{2A{{\in }_{o}}}\] |
\[F=QE\] |
\[F=\frac{{{Q}^{2}}}{2A{{\in }_{o}}}\] |
But \[Q=CV=\frac{{{\in }_{o}}A(V)}{d}\] |
\[\therefore F=\frac{{{({{\in }_{o}}AV)}^{2}}}{{{d}^{2}}\times 2A{{\in }_{o}}}\] |
\[=\frac{{{({{\in }_{o}}A)}^{2}}\times {{V}^{2}}}{{{d}^{2}}\times 2\times (A{{\in }_{o}})}\] |
\[=\frac{({{\in }_{o}}A){{V}^{2}}}{{{d}^{2}}\times 2}\] |
\[25\times {{10}^{-6}}=\frac{(8.85\times {{10}^{-12}})\times (200\times {{10}^{-4}})\times {{V}^{2}}}{2.25\times {{10}^{-4}}\times 2}\] |
\[V=\sqrt{\frac{25\times {{10}^{-6}}\times 2.25\times {{10}^{-4}}\times 2}{8.85\times {{10}^{-12}}\times 200\times {{10}^{-4}}}}\] |
Here, on solving, \[v\approx 250V\] |
You need to login to perform this action.
You will be redirected in
3 sec