(B) For the particle to be in equilibrium,the electric force must balance the gravitational force:$QE = Mg$Since $Q = Ne$ (where $N$ is an integer) an

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(B) For the particle to be in equilibrium,the electric force must balance the gravitational force:$QE = Mg$Since $Q = Ne$ (where $N$ is an integer) and $E = V/d$,we have:$(Ne) \left( \frac{V}{d} \right) = Mg$$N = \frac{Mgd}{eV}$Given $M = 1.96 \times 10^{-15} \, kg$,$g = 10 \, m/s^2$,$d = 0.02 \, m$,$V = 800 \, V$,and $e = 1.6 \times 10^{-19} \, C$:$N = \frac{(1.96 \times 10^{-15}) \times 10 \times 0.02}{(1.6 \times 10^{-19}) \times 800}$$N = \frac{3.92 \times 10^{-16}}{1.28 \times 10^{-16}} = 3$Therefore,the charge on the particle is $Q = 3e$.

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

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The potential difference between two metal plates is $800 \, V$ and they are separated by a horizontal distance of $0.02 \, m$. $A$ particle of mass $1.96 \times 10^{-15} \, kg$ is held in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is .......

A
$e$
B
$3e$
C
$6e$
D
$8e$

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Electric Charges and Fields

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Motion of Charge particle in Electric filed

$A$ proton, a deuteron, and an $\alpha$-particle having the same momentum enter a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

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$A$ positively charged particle moving along the horizontal axis with a certain velocity enters a uniform electric field directed along the vertical axis. Its

Easy

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$A$ uniform electric field of $10^3 \, V/m$ is directed along the $y$-axis. $A$ particle of mass $1 \, g$ and charge $10^{-6} \, C$ is projected from the origin into the field with a velocity of $10 \, m/s$ along the positive $x$-axis. Its speed after $10 \, s$ is (in $m/s$):

Medium

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Kinetic energy of an electron accelerated in a potential difference of $100 \, V$ is

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The electric field at a point is given by $\vec E = E_0 \hat i \, V/m$. $A$ particle of charge $+q_0$ moves from point $A(0, a)$ to $B(a, 0)$ along a circular path as shown in the figure. Find the work done by the electric field in this motion.

Medium

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The potential difference between two metal plates is $800 \, V$ and they are separated by a horizontal distance of $0.02 \, m$. $A$ particle of mass $1.96 \times 10^{-15} \, kg$ is held in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is .......

Similar Questions

$A$ proton, a deuteron, and an $\alpha$-particle having the same momentum enter a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

$A$ positively charged particle moving along the horizontal axis with a certain velocity enters a uniform electric field directed along the vertical axis. Its

$A$ uniform electric field of $10^3 \, V/m$ is directed along the $y$-axis. $A$ particle of mass $1 \, g$ and charge $10^{-6} \, C$ is projected from the origin into the field with a velocity of $10 \, m/s$ along the positive $x$-axis. Its speed after $10 \, s$ is (in $m/s$):

Kinetic energy of an electron accelerated in a potential difference of $100 \, V$ is

The electric field at a point is given by $\vec E = E_0 \hat i \, V/m$. $A$ particle of charge $+q_0$ moves from point $A(0, a)$ to $B(a, 0)$ along a circular path as shown in the figure. Find the work done by the electric field in this motion.

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