Motion of Charge particle in Electric filed Questions in English

(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 $\vec{E} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ N/C}$.
$W = 3 \times (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (5\hat{i} + 3\hat{j} + 7\hat{k})$.
$W = 3 \times [(2 \times 5) + (3 \times 3) + (4 \times 7)]$.
$W = 3 \times [10 + 9 + 28] = 3 \times 47 = 141 \text{ J}$.

(B) $1$. The work done by the uniform electric field '$E$' on the charge '$q$' over a distance '$D$' is equal to the kinetic energy gained by the particle: $W = F \cdot D = qED$.
$2$. Since the particle starts from rest, its kinetic energy at the end of distance '$D$' is $K = qED$.
$3$. When the particle approaches the fixed charge '$Q$',it experiences a repulsive force. At the point of closest approach '$r$',the kinetic energy of the particle is completely converted into electrostatic potential energy.
$4$. The electrostatic potential energy at distance '$r$' is given by $U = \frac{1}{4\pi\epsilon_0} \frac{qQ}{r}$.
$5$. By the law of conservation of energy: $qED = \frac{1}{4\pi\epsilon_0} \frac{qQ}{r}$.
$6$. Solving for '$r$': $r = \frac{Q}{4\pi\epsilon_0ED}$.

(A) $1$. The force on the electron is $F = eE = (1.6 \times 10^{-19} \text{ C})(2.0 \times 10^4 \text{ N/C}) = 3.2 \times 10^{-15} \text{ N}$.
$2$. The acceleration of the electron is $a = \frac{F}{m_e} = \frac{3.2 \times 10^{-15} \text{ N}}{9.11 \times 10^{-31} \text{ kg}} \approx 3.51 \times 10^{15} \text{ m/s}^2$.
$3$. Using the equation of motion $s = ut + \frac{1}{2}at^2$, where $u = 0$ and $s = 1.5 \text{ cm} = 0.015 \text{ m}$:
$0.015 = 0 + \frac{1}{2}(3.51 \times 10^{15})t^2$.
$4$. $t^2 = \frac{2 \times 0.015}{3.51 \times 10^{15}} = \frac{0.03}{3.51 \times 10^{15}} \approx 8.55 \times 10^{-18} \text{ s}^2$.
$5$. $t = \sqrt{8.55 \times 10^{-18}} \approx 2.92 \times 10^{-9} \text{ s}$.