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Two plates are 5 cm apart, a potential difference of 10V is applied between them. What is the electric field between the plates?

$\begin{array}{1 1} 20 \;N/C \\ 200 \;N/C \\ 50 \;N/C \\ 500 \;N/C \end{array}$

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  • The Electric field between two plates is given by $E = \large\frac{V}{d}$, where V is the potential difference in volts and d is the distance between the plates.
Given $V = 10V$ and $d = 5 cm$, $E = \large\frac{V}{d} $$=\large \frac{10}{5\times10^{-2}}$ = 200 N/C

 

answered Dec 20, 2013 by balaji.thirumalai
 

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