Motion of Charge particle in Electric filed Questions in English

(D) The particle moves with constant velocity $u$ along the $X$-axis, so $x = ut$, which implies $t = x/u$.
Along the $Y$-axis, the force on the particle is $F = eE$, so the acceleration is $a_y = \frac{eE}{m}$.
The displacement along the $Y$-axis is $y = \frac{1}{2} a_y t^2 = \frac{1}{2} \left( \frac{eE}{m} \right) \left( \frac{x}{u} \right)^2 = \left( \frac{eE}{2mu^2} \right) x^2$.
This is the equation of a parabola of the form $x^2 = 4ay$.
Rearranging the equation: $x^2 = \left( \frac{2mu^2}{eE} \right) y$.
Comparing this with the standard form $x^2 = 4ay$, we get $4a = \frac{2mu^2}{eE}$.
Therefore, the focal length $a = \frac{2mu^2}{4eE} = \frac{mu^2}{2eE}$.

(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.

(B) The particle is accelerated by a uniform electric field $E$ over a distance $D$. The work done by the electric field is equal to the kinetic energy $(KE)$ gained by the particle.
$KE = W = q E D$
At the point of closest approach $(r_0)$,the entire kinetic energy of the particle is converted into electrostatic potential energy $(PE)$ due to the interaction with the fixed charge $Q$.
$PE = \frac{1}{4 \pi \varepsilon_0} \frac{q Q}{r_0}$
Equating $KE$ and $PE$ at the point of closest approach:
$q E D = \frac{1}{4 \pi \varepsilon_0} \frac{q Q}{r_0}$
Solving for $r_0$:
$r_0 = \frac{Q}{4 \pi \varepsilon_0 E D}$

(C) When the pendulum is in equilibrium under the influence of gravity ($mg$ acting downwards) and an electric force ($qE$ acting horizontally),the net force acting on the bob is the vector sum of these two forces.
Since the forces are perpendicular to each other,the magnitude of the net force $F_{net}$ is given by $F_{net} = \sqrt{(mg)^{2} + (qE)^{2}}$.
At equilibrium,the tension $T$ in the string must balance this net force to keep the bob stationary relative to the point of suspension.
Therefore,the tension $T = \sqrt{(mg)^{2} + (qE)^{2}}$.

(C) The force on the charge is $F = qE$.
Using Newton's second law,the acceleration $a$ is given by $a = \frac{F}{M} = \frac{qE}{2m}$.
Starting from rest $(u = 0)$,the speed $v$ after $n$ seconds is calculated using the equation of motion $v = u + at$.
Substituting the values,we get $v = 0 + (\frac{qE}{2m}) \times n = \frac{qEn}{2m}$.
Therefore,the correct option is $C$.

(C) The force on a charged particle in an electric field is given by $F = qE$.
According to Newton's second law of motion,$F = ma$,therefore $a = \frac{qE}{m}$.
For a proton,the charge is $q_p = e$ and the mass is $m_p$. Thus,$a_p = \frac{eE}{m_p}$.
For an $\alpha$-particle,the charge is $q_\alpha = 2e$ and the mass is $m_\alpha = 4m_p$.
Substituting these values,$a_\alpha = \frac{2eE}{4m_p} = \frac{1}{2} \frac{eE}{m_p} = \frac{a_p}{2}$.
Therefore,the ratio $\frac{a_p}{a_\alpha} = 2$,which means $a_p : a_\alpha = 2 : 1$.

(A) The change in kinetic energy $(\Delta K)$ is equal to the total work done by all forces acting on the particle.
Since the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity $\vec{v}$,it does no work on the particle.
Therefore,the work is done only by the electric force $\vec{F}_e = q\vec{E} = qE_0 e^{-\lambda x}\hat{i}$.
The work done $W$ as the particle moves from $x = 0$ to $x = L$ is given by:
$W = \int_{0}^{L} F_x dx = \int_{0}^{L} q E_0 e^{-\lambda x} dx$
$W = q E_0 \left[ \frac{e^{-\lambda x}}{-\lambda} \right]_{0}^{L}$
$W = \frac{q E_0}{-\lambda} (e^{-\lambda L} - e^0) = \frac{q E_0}{\lambda} (1 - e^{-\lambda L})$.
Thus,the change in kinetic energy is $\frac{q E_0}{\lambda} (1 - e^{-\lambda L})$.

(D) The work done $W$ in moving a charge $q$ in a uniform electric field $\vec{E}$ is given by $W = q \vec{E} \cdot \vec{d}$,where $\vec{d}$ is the displacement vector.
Given $q = 3 \text{ C}$.
The displacement vector $\vec{d} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.
$\vec{d} = (5 - 0)\hat{i} + (1 - (-2))\hat{j} + (2 - (-5))\hat{k} = 5\hat{i} + 3\hat{j} + 7\hat{k}$.
Given $\vec{E} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ N/C}$.
$W = 3 \times (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (5\hat{i} + 3\hat{j} + 7\hat{k})$.
$W = 3 \times [(2 \times 5) + (3 \times 3) + (4 \times 7)]$.
$W = 3 \times [10 + 9 + 28] = 3 \times 47 = 141 \text{ J}$.