Two plates are $20 c m$ apart and the potential difference between them is $10 V$ . The electric field between the plates is

50 V m^{- 1}

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500 V m^{- 1}

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0.5 V m^{- 1}

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20 V m^{- 1}

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Solution

The correct option is D $50 V m^{- 1}$
Potential gradient relates with electric field according to the relation, $E = - \frac{d V}{d r}$
$= - \frac{10}{20 \times 10^{- 2}}$
$= 50 V m^{- 1}$ .

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Two plates are 20 cm apart and the potential difference between them is 10 V. The electric field between the plates is

The correct option is D 50 Vm−1Potential gradient relates with electric field according to the relation, E=−dVdr=−1020×10−2=50 Vm−1.

Two plates are $20 c m$ apart and the potential difference between them is $10 V$ . The electric field between the plates is

The correct option is D $50 V m^{- 1}$
Potential gradient relates with electric field according to the relation, $E = - \frac{d V}{d r}$
$= - \frac{10}{20 \times 10^{- 2}}$
$= 50 V m^{- 1}$ .