In moving from A to B along an electric field line, electric field does 6.4 × {10^{

English

Gujarati

Hindi

2. Electric Potential and Capacitance

medium

In moving from $A$ to $B$ along an electric field line, the electric field does $6.4 \times {10^{ - 19}}\,J$ of work on an electron. If ${\phi _1},\;{\phi _2}$ are equipotential surfaces, then the potential difference $({V_C} - {V_A})$ is.....$V$

$-4$

$4$

$0$

$64$

Solution

(b) Work done by the field $W = q( – dV) = – e({V_A} – {V_B})$
= $e({V_B} – {V_A})$ = $e({V_C} – {V_A})$ $(\because V_B = V_C)$
$==>$ $({V_C} – {V_A})$ $ = \frac{W}{e} = \frac{{6.4 \times {{10}^{ – 19}}}}{{1.6 \times {{10}^{ – 19}}}} = 4\,V$

Standard 12

Physics

A two point charges $4 q$ and $-q$ are fixed on the $x-$axis at $x=-\frac{d}{2}$ and $x=\frac{d}{2},$ respectively. If a third point charge $'q'$ is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will

hard

(JEE MAIN-2020)

There is an electric field $E$ in $X$-direction. If the work done on moving a charge $0.2\,C$ through a distance of $2\,m$ along a line making an angle $60^\circ $ with the $X$-axis is $4.0\;J$, what is the value of $E$…….. $N/C$

medium

(AIPMT-1995)

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $V$ in vacuum. The final speed of the electron will be

easy

The charge $q$ is fired towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$, the closest distance of approach would be

easy

An electron enters in high potential region ${V_2}$ from lower potential region ${V_1}$ then its velocity

easy

In moving from $A$ to $B$ along an electric field line, the electric field does $6.4 \times {10^{ - 19}}\,J$ of work on an electron. If ${\phi _1},\;{\phi _2}$ are equipotential surfaces, then the potential difference $({V_C} - {V_A})$ is.....$V$

Solution

Similar Questions

A two point charges $4 q$ and $-q$ are fixed on the $x-$axis at $x=-\frac{d}{2}$ and $x=\frac{d}{2},$ respectively. If a third point charge $'q'$ is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will

There is an electric field $E$ in $X$-direction. If the work done on moving a charge $0.2\,C$ through a distance of $2\,m$ along a line making an angle $60^\circ $ with the $X$-axis is $4.0\;J$, what is the value of $E$…….. $N/C$

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $V$ in vacuum. The final speed of the electron will be

The charge $q$ is fired towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$, the closest distance of approach would be

An electron enters in high potential region ${V_2}$ from lower potential region ${V_1}$ then its velocity

Start a Free Trial Now