(D) When the electric field $E$ is switched on,the block experiences an electric force $F_e = qE$ in the direction opposite to the electric field (sin

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(D) When the electric field $E$ is switched on,the block experiences an electric force $F_e = qE$ in the direction opposite to the electric field (since the charge is negative). As the block moves,the spring is compressed. At maximum compression $x$,the velocity of the block becomes zero. According to the work-energy theorem,the work done by the electric force is equal to the potential energy stored in the spring: $W_{electric} = U_{spring}$ $F_e \cdot x = \frac{1}{2} k x^2$ $(qE) x = \frac{1}{2} k x^2$ Solving for $x$ (where $x \neq 0$): $qE = \frac{1}{2} k x$ $x = \frac{2qE}{k}$

Class 12 Physics Electric Charges and Fields Motion of Charge particle in Electric filed

Medium

$A$ block of mass $m$ containing a net negative charge $-q$ is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring constant $k$ as shown. If a horizontal electric field $E$ parallel to the spring is switched on,then the maximum compression of the spring is:

A
$\sqrt{qE/k}$
B
$\sqrt{k/qE}$
C
$qE/k$
D
$\frac{2qE}{k}$

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Electric Charges and Fields

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Motion of Charge particle in Electric filed

$A$ particle of mass $m$ and charge $q$ is released from rest in a uniform electric field. If there is no other force on the particle,the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)

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$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

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An $\alpha$-particle of mass $6.4 \times 10^{-27} \ kg$ and charge $3.2 \times 10^{-19} \ C$ is situated in a uniform electric field of $1.6 \times 10^{5} \ Vm^{-1}$. The velocity of the particle at the end of $2 \times 10^{-2} \ m$ path when it starts from rest is

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As shown in the figure, an electron is projected with kinetic energy $K$ at an angle of $\theta = 45^\circ$ between two charged plates. The electron strikes the upper plate. The value of the electric field $E$ must be greater than:

Medium

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$A$ charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E$. Neglecting the effect of gravity,the kinetic energy of the charged particle after $t$ seconds is

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$A$ particle of mass $m$ and charge $q$ is released from rest in a uniform electric field. If there is no other force on the particle,the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)

$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

An $\alpha$-particle of mass $6.4 \times 10^{-27} \ kg$ and charge $3.2 \times 10^{-19} \ C$ is situated in a uniform electric field of $1.6 \times 10^{5} \ Vm^{-1}$. The velocity of the particle at the end of $2 \times 10^{-2} \ m$ path when it starts from rest is

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$A$ charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E$. Neglecting the effect of gravity,the kinetic energy of the charged particle after $t$ seconds is

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