(C) The electric field $E$ between two parallel plates is given by the formula $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the

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(C) The electric field $E$ between two parallel plates is given by the formula $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates.Given:Potential difference $V = 10 \ V$Distance $d = 20 \ cm = 20 \times 10^{-2} \ m = 0.2 \ m$Substituting these values into the formula:$E = \frac{10}{0.2} = 50 \ Vm^{-1}$Therefore,the electric field between the plates is $50 \ Vm^{-1}$.

Class 12 Physics Electric Potential and Capacitance Relation between Electric Field and Potential and Potential Gradient

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Two plates are $20 \ cm$ apart and a potential difference of $10 \ V$ is applied between them. The electric field between the plates is . . . . . . . (in $Vm^{-1}$)

GUJCET 2010 All GUJCET

A
$0.5$
B
$20$
C
$50$
D
$500$

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Class 12

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Chapter

Electric Potential and Capacitance

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Relation between Electric Field and Potential and Potential Gradient

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Two plates are $20 \ cm$ apart and a potential difference of $10 \ V$ is applied between them. The electric field between the plates is . . . . . . . (in $Vm^{-1}$)

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