(D) The electric field $E$ between the parallel plates is given by $E = \frac{V}{d}$.Given, $V = 10^4 \,V$ and $d = 0.5 \,cm = 0.5 \times 10^{-2} \,m$

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(D) The electric field $E$ between the parallel plates is given by $E = \frac{V}{d}$.Given, $V = 10^4 \,V$ and $d = 0.5 \,cm = 0.5 \times 10^{-2} \,m$.Therefore, $E = \frac{10^4}{0.5 \times 10^{-2}} = 2 \times 10^6 \,V/m$.The force $F$ on an electron of charge $e = 1.6 \times 10^{-19} \,C$ is given by $F = eE$.$F = (1.6 \times 10^{-19} \,C) \times (2 \times 10^6 \,V/m) = 3.2 \times 10^{-13} \,N$.

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

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The potential difference between two parallel plates is $10^4 \,V$. If the plates are separated by $0.5 \,cm$, the force on an electron between the plates is

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A
$32 \times 10^{-13} \,N$
B
$0.32 \times 10^{-13} \,N$
C
$0.032 \times 10^{-13} \,N$
D
$3.2 \times 10^{-13} \,N$

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Electric Potential and Capacitance

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Relation between Electric Field and Potential and Potential Gradient

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The potential difference between two parallel plates is $10^4 \,V$. If the plates are separated by $0.5 \,cm$, the force on an electron between the plates is

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