(A) Given: Mass of the particle $= M$, Charge $= q$, Electric field $= E$, Initial velocity $u = 0$.Acceleration $a = F/M = qE/M$.Distance travelled $

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(A) Given: Mass of the particle $= M$, Charge $= q$, Electric field $= E$, Initial velocity $u = 0$.Acceleration $a = F/M = qE/M$.Distance travelled $S = ut + (1/2)at^2 = (1/2)(qE/M)t^2$.Final velocity $v = u + at = (qE/M)t$.Kinetic energy $T = (1/2)Mv^2 = (1/2)M(qEt/M)^2 = (1/2)(q^2E^2t^2/M)$.Now, the ratio $T/S = [(1/2)(q^2E^2t^2/M)] / [(1/2)(qE/M)t^2] = qE$.Since $q$ and $E$ are constants, the ratio $T/S = qE$ is independent of time $t$ and remains constant.

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

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$A$ particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field of magnitude $E$. After a time $t$, the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then, the ratio $T/S$:

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A
remains constant with time $t$
B
varies linearly with the mass $M$ of the particle
C
is independent of the charge $q$
D
is independent of the magnitude of the electric field $E$

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Electric Charges and Fields

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Motion of Charge particle in Electric filed

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$A$ particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field of magnitude $E$. After a time $t$, the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then, the ratio $T/S$:

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Two charged particles of masses $m$ and $2m$ have charges $+2q$ and $+q$ respectively. They are kept in a uniform electric field and allowed to move for the same time. The ratio of their kinetic energies will be

An electron experiences a force equal to its weight when placed in an electric field. The intensity of the field will be

The $e/m$ ratio for an electron is $1.8 \times 10^{11} \ C \ kg^{-1}$. What will be its velocity when accelerated through a potential difference of $9 \ V$?

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