(D) The electric field $E$ between two parallel plates separated by a distance $d$ with a potential difference $V$ is given by $E = \frac{V}{d}$.The f

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(D) The electric field $E$ between two parallel plates separated by a distance $d$ with a potential difference $V$ is given by $E = \frac{V}{d}$.The force $F$ acting on a particle of charge $q$ in this electric field is $F = qE = \frac{qV}{d}$.According to Newton's second law of motion,the acceleration $a$ is given by $a = \frac{F}{m}$.Substituting the value of $F$,we get $a = \frac{qV}{md}$.

Class 12 Physics Electric Potential and Capacitance Parallel Plate Capacitor

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Two parallel plates separated by $d \ mm$ are kept at a potential difference of $V \ V$. $A$ particle of mass $m$ and charge $q$ enters the region between them with some velocity. The acceleration of the particle will be

MHT CET 2022 All MHT CET

A
$\frac{q}{dmV}$
B
$\frac{qm}{Vd}$
C
$\frac{qd}{Vm}$
D
$\frac{qV}{dm}$

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Class 12

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Electric Potential and Capacitance

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Parallel Plate Capacitor

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DifficultAP EAMCET 2023

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Medium

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Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

MediumJEE MAIN 2022

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$A$ parallel plate capacitor is to be designed,using a dielectric of dielectric constant $5$,so as to have a dielectric strength of $10^9 \; Vm^{-1}$. If the voltage rating of the capacitor is $12 \; kV$,the minimum area of each plate required to have a capacitance of $80 \; pF$ is

MediumNEET 2017

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If the charge on parallel plates is $\pm Q$ and the area is $A$,write the formula for the electric field between the two plates.

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Two parallel plates separated by $d \ mm$ are kept at a potential difference of $V \ V$. $A$ particle of mass $m$ and charge $q$ enters the region between them with some velocity. The acceleration of the particle will be

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Four metal plates,each with surface area $A$ on one side,are placed with separation '$d$' as shown in the figure. The capacitance between $a$ and $b$ is ($\varepsilon_0$ = permittivity of free space).

The unit of absolute permittivity is

Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

$A$ parallel plate capacitor is to be designed,using a dielectric of dielectric constant $5$,so as to have a dielectric strength of $10^9 \; Vm^{-1}$. If the voltage rating of the capacitor is $12 \; kV$,the minimum area of each plate required to have a capacitance of $80 \; pF$ is

If the charge on parallel plates is $\pm Q$ and the area is $A$,write the formula for the electric field between the two plates.

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