(B) Given: Distance $d = 1 \ mm = 10^{-3} \ m$, Electric field $E = 5 \times 10^5 \ V/m$, Charge of electron $q = 1.6 \times 10^{-19} \ C$.First, calc

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(B) Given: Distance $d = 1 \ mm = 10^{-3} \ m$, Electric field $E = 5 \times 10^5 \ V/m$, Charge of electron $q = 1.6 \times 10^{-19} \ C$.First, calculate the potential difference $(V)$ between the plates: $V = E \times d = (5 \times 10^5 \ V/m) \times (10^{-3} \ m) = 500 \ V$.The change in potential energy $(\Delta PE)$ is given by $\Delta PE = q \times V$.Substituting the values: $\Delta PE = (1.6 \times 10^{-19} \ C) \times (500 \ V) = 800 \times 10^{-19} \ J = 8 \times 10^{-17} \ J$.

Class 12 Physics Electric Potential and Capacitance Potential Energy and Work Done in uniform and non uniform Electric Field

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The distance between the plates of a capacitor is $1 \ mm$. An electric field of $5 \times 10^5 \ V/m$ is produced between the two plates. What will be the change in the potential energy $(PE)$ of an electron if it is moved from one plate to the other?

A
$5 \times 10^{-19} \ J$
B
$8 \times 10^{-17} \ J$
C
$5 \times 10^{-17} \ J$
D
$5 \times 10^2 \ J$

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Class 12

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Chapter

Electric Potential and Capacitance

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Potential Energy and Work Done in uniform and non uniform Electric Field

$A$ charge of $2 \ \mu C$ moves from point $B$ to point $C$ along the path shown in the figure. Calculate the work done in $J$.

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$A$ particle of mass $100\, g$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force. (Result in $m/s$) (in $.73$)

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Two positive point charges of $12\,\mu C$ and $8\,\mu C$ are $10\,cm$ apart. The work done in bringing them $4\,cm$ closer is

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$A$ charge $(-q)$ and another charge $(+Q)$ are kept at two points $A$ and $B$ respectively. Keeping the charge $(+Q)$ fixed at $B$,the charge $(-q)$ at $A$ is moved to another point $C$ such that $ABC$ forms an equilateral triangle of side $\ell$. The net work done in moving the charge $(-q)$ is

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$A$ region contains a uniform electric field $E = (10 \hat{i} + 30 \hat{j}) \ Vm^{-1}$. $A$ and $B$ are two points in the field at $(1, 2, 0) \ m$ and $(2, 1, 3) \ m$,respectively. The work done when a charge of $0.8 \ C$ moves from $A$ to $B$ in a parabolic path is (in $J$)

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The distance between the plates of a capacitor is $1 \ mm$. An electric field of $5 \times 10^5 \ V/m$ is produced between the two plates. What will be the change in the potential energy $(PE)$ of an electron if it is moved from one plate to the other?

Similar Questions

$A$ charge of $2 \ \mu C$ moves from point $B$ to point $C$ along the path shown in the figure. Calculate the work done in $J$.

$A$ particle of mass $100\, g$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force. (Result in $m/s$) (in $.73$)

Two positive point charges of $12\,\mu C$ and $8\,\mu C$ are $10\,cm$ apart. The work done in bringing them $4\,cm$ closer is

$A$ charge $(-q)$ and another charge $(+Q)$ are kept at two points $A$ and $B$ respectively. Keeping the charge $(+Q)$ fixed at $B$,the charge $(-q)$ at $A$ is moved to another point $C$ such that $ABC$ forms an equilateral triangle of side $\ell$. The net work done in moving the charge $(-q)$ is

$A$ region contains a uniform electric field $E = (10 \hat{i} + 30 \hat{j}) \ Vm^{-1}$. $A$ and $B$ are two points in the field at $(1, 2, 0) \ m$ and $(2, 1, 3) \ m$,respectively. The work done when a charge of $0.8 \ C$ moves from $A$ to $B$ in a parabolic path is (in $J$)

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