Motion of Charge particle in Electric filed Questions in English

(B) We know that the force on the electron is $F = qE$ and from Newton's second law,$F = ma$.
Equating these,we get $ma = qE$,which implies the acceleration $a = \frac{qE}{m}$.
Using the third equation of motion,$v^2 - u^2 = 2aS$,where $u = 0$ (starts from rest) and $S = L$:
$v^2 - 0^2 = 2 \left( \frac{qE}{m} \right) L$
$v^2 = \frac{2qEL}{m}$
$v = \sqrt{\frac{2qEL}{m}}$

(B) The force acting on the charged particle in a uniform electric field $E$ is given by $F = qE$.
According to Newton's second law,the acceleration $a$ of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.
Since the particle starts from rest,its initial velocity $u = 0$. After time $t$,the velocity $v$ of the particle is given by $v = u + at = 0 + (\frac{qE}{m})t = \frac{qEt}{m}$.
The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.
Substituting the value of $v$,we get $K = \frac{1}{2}m(\frac{qEt}{m})^2 = \frac{1}{2}m(\frac{q^2E^2t^2}{m^2}) = \frac{q^2E^2t^2}{2m}$.

(B) The electrostatic force acting on a charged particle in a uniform electric field $E$ is given by $F = qE$.
For an electron,the magnitude of charge is $e$,so the force is $F_e = eE$. The acceleration is $a_e = \frac{F_e}{m_e} = \frac{eE}{m_e}$.
For a proton,the magnitude of charge is also $e$,so the force is $F_p = eE$. The acceleration is $a_p = \frac{F_p}{m_p} = \frac{eE}{m_p}$.
Taking the ratio of the acceleration of the electron to the acceleration of the proton:
$\frac{a_e}{a_p} = \frac{eE / m_e}{eE / m_p} = \frac{m_p}{m_e}$.

(C) The force acting on the electron in a uniform electric field is given by $F = qE$.
Since the electron is moving in the direction of the force,the work done by the electric field over a distance $L$ is $W = F \times L = qEL$.
According to the work-energy theorem,the work done is equal to the change in kinetic energy of the electron.
Since the electron starts from rest,the initial kinetic energy is $0$.
Therefore,$\frac{1}{2} mv^2 = qEL$.
Solving for the velocity $v$,we get $v^2 = \frac{2qEL}{m}$,which implies $v = \sqrt{\frac{2qEL}{m}}$.

(A) We know that the force acting on the electron is $F = ma$ and $F = qE$.
Equating these,we get $qE = ma$,which implies $a = \frac{qE}{m} \quad ...(i)$.
Using the equation of motion $v^2 - u^2 = 2as$,where $u = 0$,$s = l$,and $v$ is the final velocity:
$v^2 - 0^2 = 2al$
$v^2 = 2al$
$a = \frac{v^2}{2l} \quad ...(ii)$.
Equating $(i)$ and $(ii)$:
$\frac{qE}{m} = \frac{v^2}{2l}$.
Rearranging for the ratio $\frac{q}{m}$:
$\frac{q}{m} = \frac{v^2}{2El}$.

(B) The force $F$ acting on a charge $q$ in an electric field $E$ is given by $F = qE$.
Given charge $q = 2e$ and electric field $E$,the force is $F = 2eE$.
The acceleration $a$ is given by Newton's second law,$a = \frac{F}{m_{total}}$.
Given mass $m_{total} = 4m$,the acceleration is $a = \frac{2eE}{4m}$.
Simplifying the expression,we get $a = \frac{eE}{2m}$.

(B) The force $F$ experienced by an electron of charge $e$ in an electric field $E$ is given by $F = eE$.
According to Newton's second law of motion,$F = ma$,where $m_e$ is the mass of the electron and $a$ is its acceleration.
Equating the two expressions for force: $ma = eE$.
Therefore,the acceleration $a$ is given by $a = \frac{eE}{m_e}$.
Given:
Charge of electron $e = 1.6 \times 10^{-19} \ C$
Electric field $E = 2.0 \times 10^4 \ NC^{-1}$
Mass of electron $m_e = 9.1 \times 10^{-31} \ kg$
Substituting the values:
$a = \frac{(1.6 \times 10^{-19} \ C) \times (2.0 \times 10^4 \ NC^{-1})}{9.1 \times 10^{-31} \ kg}$
$a = \frac{3.2 \times 10^{-15}}{9.1 \times 10^{-31}} \ ms^{-2}$
$a \approx 0.3516 \times 10^{16} \ ms^{-2}$
$a \approx 3.52 \times 10^{15} \ ms^{-2}$.

(D) The force acting on the particle due to the electric field is given by $F = qE$.
Since the particle starts from rest and moves in the direction of the force,the work done by the electric field on the particle over a distance $x$ is $W = F \cdot x = (qE) \cdot x = qEx$.
According to the work-energy theorem,the work done on a particle is equal to the change in its kinetic energy.
Since the initial kinetic energy is $0$,the final kinetic energy is equal to the work done,which is $qEx$.

(A) For a charge $q_2$ to revolve in a circular orbit,the electrostatic force between $q_1$ and $q_2$ provides the necessary centripetal force.
The centripetal force is given by $F_c = \frac{m v^2}{r}$,where $v$ is the orbital velocity.
The electrostatic force is given by Coulomb's law as $F_e = \frac{k q_1 q_2}{r^2}$.
Equating the two forces: $\frac{m v^2}{r} = \frac{k q_1 q_2}{r^2}$.
Since $v = r \omega$,where $\omega$ is the angular velocity,we have $\frac{m (r \omega)^2}{r} = \frac{k q_1 q_2}{r^2}$.
Simplifying,$m r \omega^2 = \frac{k q_1 q_2}{r^2}$,which gives $\omega^2 = \frac{k q_1 q_2}{m r^3}$.
Since $\omega = \frac{2 \pi}{T}$,where $T$ is the periodic time,we have $\left(\frac{2 \pi}{T}\right)^2 = \frac{k q_1 q_2}{m r^3}$.
$\frac{4 \pi^2}{T^2} = \frac{k q_1 q_2}{m r^3}$.
Therefore,$T^2 = \frac{4 \pi^2 m r^3}{k q_1 q_2}$.
Taking the square root,$T = \left[\frac{4 \pi^2 m r^3}{k q_1 q_2}\right]^{\frac{1}{2}}$.

(C) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K.E. = eV$.
Since the electron starts from rest, its kinetic energy is $\frac{1}{2}mv^2$.
Equating the two, we get $\frac{1}{2}mv^2 = eV$.
Solving for velocity $v$, we have $v = \sqrt{\frac{2eV}{m}}$.
Substituting the values: $e = 1.6 \times 10^{-19} \,C$, $m = 9.1 \times 10^{-31} \,kg$, and $V = 1200 \,V$.
$v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1200}{9.1 \times 10^{-31}}} = \sqrt{\frac{3.84 \times 10^{-16}}{9.1 \times 10^{-31}}} = \sqrt{0.42198 \times 10^{15}} = \sqrt{42.198 \times 10^{13}} \approx 2.05 \times 10^{7} \,ms^{-1}$.
Rounding this value, we get $v \approx 2.1 \times 10^{7} \,ms^{-1}$.

(A) Given: mass of electron $= m$,charge $= e$,distance $= h$,electric field $= E$.
The force on the electron in the electric field is $F = eE$.
Using Newton's second law,the acceleration $a$ is given by $a = \frac{F}{m} = \frac{eE}{m}$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$,where initial velocity $u = 0$ and $S = h$:
$h = 0 + \frac{1}{2} \left( \frac{eE}{m} \right) t^2$.
Rearranging for $t^2$:
$t^2 = \frac{2hm}{eE}$.
Taking the square root on both sides:
$t = \sqrt{\frac{2hm}{eE}}$.

(C) The force acting on a charged particle in a uniform electric field is given by $F = qE$.
According to Newton's second law,the acceleration $a$ of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.
Since the particle is released from rest,its initial velocity $u = 0$. The velocity $v$ after time $t$ is given by the equation $v = u + at = 0 + \frac{qE}{m}t = \frac{qEt}{m}$.
The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.
Substituting the value of $v$,we get $K = \frac{1}{2}m\left(\frac{qEt}{m}\right)^2 = \frac{1}{2}m \cdot \frac{q^2 E^2 t^2}{m^2} = \frac{E^2 q^2 t^2}{2m}$.

(C) Given: Mass $m = 2 \ g = 2 \times 10^{-3} \ kg$.
Electric field $E = 300 \ NC^{-1}$.
Initial kinetic energy $(KE)_1 = 0$ at $x = 0$.
Final kinetic energy $(KE)_2 = 0.12 \ J$ at $x = 0.5 \ m$.
According to the work-energy theorem,the work done by the electric force is equal to the change in kinetic energy.
Work done $W = F \cdot d = (QE) \cdot d$.
Change in kinetic energy $\Delta KE = KE_2 - KE_1 = 0.12 \ J - 0 \ J = 0.12 \ J$.
Equating the two: $Q \times 300 \times 0.5 = 0.12$.
$Q \times 150 = 0.12$.
$Q = \frac{0.12}{150} = 0.0008 \ C$.
$Q = 800 \times 10^{-6} \ C = 800 \ \mu C$.
Since the object gains kinetic energy moving in the direction of the electric field,the charge $Q$ must be positive.

(C) Since the charges are equal in magnitude and placed in a uniform electric field $E$,the magnitude of the force acting on each charge is $F = qE$.
Since the magnitude of the force $F$ is the same for both charges,we have $F_{1} = F_{2}$.
Using Newton's second law,$F = ma$,we can write $m_{1}a_{1} = m_{2}a_{2}$.
Rearranging this gives the ratio of accelerations as $\frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}}$.
Given that $\frac{m_{1}}{m_{2}} = 0.5$,we have $\frac{m_{2}}{m_{1}} = \frac{1}{0.5} = 2$.
Therefore,the ratio of their accelerations is $\frac{a_{1}}{a_{2}} = 2$.

(A) According to the work-energy theorem,the work done by the electric field is equal to the change in the kinetic energy of the electron.
Work done $W = qEx = \frac{1}{2}mv^2$.
Rearranging for velocity $v$,we get $v = \sqrt{\frac{2qEx}{m}} = \sqrt{2 \left(\frac{e}{m}\right) Ex}$.
Given values: $\frac{e}{m} = 1.8 \times 10^{11} \text{ C kg}^{-1}$,$E = 1 \times 10^{4} \text{ N C}^{-1}$,and $x = 2 \times 10^{-2} \text{ m}$.
Substituting these values:
$v = \sqrt{2 \times (1.8 \times 10^{11}) \times (1 \times 10^{4}) \times (2 \times 10^{-2})}$
$v = \sqrt{7.2 \times 10^{13}} = \sqrt{72 \times 10^{12}}$
$v \approx 8.485 \times 10^{6} \text{ m s}^{-1} \approx 8.5 \times 10^{6} \text{ m s}^{-1}$.

(B) The force $F$ experienced by a charged particle with charge $q$ in a uniform electric field $E$ is given by $F = qE$.
According to Newton's second law of motion,the force is also given by $F = ma$,where $m$ is the mass and $a$ is the acceleration.
Equating the two expressions for force: $ma = qE$.
Therefore,the acceleration $a$ is given by $a = \frac{qE}{m}$.

(D) Given: Mass of $\alpha$-particle $m = 6.4 \times 10^{-27} \ kg$,charge $q = 3.2 \times 10^{-19} \ C$,electric field $E = 1.6 \times 10^{5} \ Vm^{-1}$,and distance $s = 2 \times 10^{-2} \ m$.
The force acting on the $\alpha$-particle is $F = qE = (3.2 \times 10^{-19}) \times (1.6 \times 10^{5}) = 5.12 \times 10^{-14} \ N$.
The acceleration $a$ of the particle is $a = \frac{F}{m} = \frac{5.12 \times 10^{-14}}{6.4 \times 10^{-27}} = 0.8 \times 10^{13} \ ms^{-2} = 8 \times 10^{12} \ ms^{-2}$.
Using the equation of motion $v^2 = u^2 + 2as$,where initial velocity $u = 0$:
$v^2 = 0 + 2 \times (8 \times 10^{12}) \times (2 \times 10^{-2}) = 32 \times 10^{10}$.
Taking the square root,$v = \sqrt{32 \times 10^{10}} = \sqrt{16 \times 2} \times 10^{5} = 4 \sqrt{2} \times 10^{5} \ ms^{-1}$.

(B) Given that the oil drop is at rest,the gravitational force acting downwards must be balanced by the electric force acting upwards.
$qE = mg$
Since the electric field $E$ between two plates is given by $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates:
$q \left(\frac{V}{d}\right) = mg$
Rearranging to solve for the charge $q$:
$q = \frac{mgd}{V}$
Given values:
$m = 10^{-6} \,kg$
$g = 10 \,ms^{-2}$
$d = 1 \,mm = 10^{-3} \,m$
$V = 500 \,V$
Substituting these values into the equation:
$q = \frac{10^{-6} \times 10 \times 10^{-3}}{500}$
$q = \frac{10^{-8}}{500} = \frac{10^{-8}}{5 \times 10^2} = 0.2 \times 10^{-10} \,C = 2 \times 10^{-11} \,C$

(B) Given: Mass $m = 0.2 \ g = 0.2 \times 10^{-3} \ kg$,Charge $q = 2 \ C$,Electric field $E = 20 \ N \ C^{-1}$,Distance $d = 20 \ cm = 0.2 \ m$,Initial velocity $u = 0$.
The force acting on the particle is $F = qE = 2 \times 20 = 40 \ N$.
According to the work-energy theorem,the work done by the electric force is equal to the change in kinetic energy.
Work done $W = F \times d = 40 \ N \times 0.2 \ m = 8 \ J$.
Since the particle starts from rest,the initial kinetic energy is $0$.
Therefore,the final kinetic energy is $K.E. = W = 8 \ J$.

(A) Given: Mass of deuteron $m = 3.2 \times 10^{-27} \ kg$,Charge of deuteron $q = e = 1.6 \times 10^{-19} \ C$,Acceleration due to gravity $g = 9.8 \ m/s^2$.
For the deuteron to be suspended freely in air,the upward electric force must balance the downward gravitational force.
Equating the forces: $qE = mg$.
Rearranging for the electric field: $E = \frac{mg}{q}$.
Substituting the values: $E = \frac{3.2 \times 10^{-27} \times 9.8}{1.6 \times 10^{-19}}$.
$E = 2 \times 9.8 \times 10^{-27+19} \ NC^{-1}$.
$E = 19.6 \times 10^{-8} \ NC^{-1}$.

(D) Given:
Mass $m = 0.5 \ g = 0.5 \times 10^{-3} \ kg$
Charge $q = 10 \ \mu C = 10 \times 10^{-6} \ C$
Electric field $E = 8 \ NC^{-1}$
Initial velocity $u = 0 \ ms^{-1}$
Time $t = 5 \ s$
The force acting on the particle is $F = qE$.
According to Newton's second law,$F = ma$,so the acceleration $a = \frac{qE}{m}$.
Substituting the values:
$a = \frac{10 \times 10^{-6} \times 8}{0.5 \times 10^{-3}} = \frac{80 \times 10^{-6}}{0.5 \times 10^{-3}} = 160 \times 10^{-3} = 0.16 \ ms^{-2}$.
Using the first equation of motion $v = u + at$:
$v = 0 + (0.16 \times 5) = 0.8 \ ms^{-1}$.

(A, B) Mass of the cork ball,$m=1 \text{ g}=10^{-3} \text{ kg}$.
Electric field,$E=(3 \hat{i}+5 \hat{j}) \times 10^5 \text{ NC}^{-1}$.
Angle,$\theta=37^{\circ}$.
Force on the cork ball due to the electric field $E$ is $F=qE$.
According to the given diagram,resolving all forces in the horizontal and vertical directions:
$T \sin \theta = q E_x \quad \dots (i)$
$T \cos \theta + q E_y = mg \implies T \cos \theta = mg - q E_y \quad \dots (ii)$
Dividing Eq. $(i)$ by Eq. $(ii)$:
$\tan \theta = \frac{q E_x}{mg - q E_y}$
Substituting the values $(\tan 37^{\circ} = 3/4)$:
$\frac{3}{4} = \frac{q \times 3 \times 10^5}{10^{-3} \times 10 - q \times 5 \times 10^5}$
$\frac{3}{4} = \frac{3q \times 10^5}{10^{-2} - 5q \times 10^5}$
$3(10^{-2} - 5q \times 10^5) = 12q \times 10^5$
$0.03 - 15q \times 10^5 = 12q \times 10^5$
$0.03 = 27q \times 10^5 \implies q = \frac{0.03}{27 \times 10^5} = \frac{1}{9} \times 10^{-7} \approx 1.11 \times 10^{-8} \text{ C}$.
Thus,$q \approx 11 \times 10^{-9} \text{ C}$ or $1.1 \times 10^{-8} \text{ C}$.
From Eq. $(i)$:
$T \sin 37^{\circ} = q E_x$
$T \times 0.6 = (1.11 \times 10^{-8}) \times (3 \times 10^5)$
$T \times 0.6 = 3.33 \times 10^{-3}$
$T = \frac{3.33 \times 10^{-3}}{0.6} = 5.55 \times 10^{-3} \text{ N}$.
Therefore,options $A$ and $B$ are correct.

(D) The electron is projected in a uniform electric field directed from $A$ to $B$. Since the electron is negatively charged,it experiences a force $F = q_e E$ in the direction opposite to the electric field,i.e.,towards plate $A$. However,the problem states the electron is projected from $A$ and we want to ensure it does not hit $B$. This implies the electric field must be directed from $B$ to $A$ or the electron is positively charged. Given the context of the trajectory shown,the electron experiences a downward acceleration $a = \frac{q_e E}{m_e}$.
For the electron not to hit plate $B$,its maximum vertical displacement $h_{\max}$ must be less than or equal to the separation distance $d = 4 \ cm = 0.04 \ m$.
The vertical component of initial velocity is $u_y = v \sin 30^{\circ} = \frac{v}{2}$.
The acceleration is $a = \frac{q_e E}{m_e} = \frac{1.6 \times 10^{-19} \times 45.5}{9.1 \times 10^{-31}} = \frac{72.8 \times 10^{-19}}{9.1 \times 10^{-31}} = 8 \times 10^{12} \ m \ s^{-2}$.
Using the kinematic equation $v_y^2 = u_y^2 - 2ah$,at maximum height $v_y = 0$:
$0 = (\frac{v}{2})^2 - 2ah_{\max} \implies h_{\max} = \frac{v^2}{8a}$.
Setting $h_{\max} = 0.04 \ m$:
$0.04 = \frac{v^2}{8 \times 8 \times 10^{12}}$
$v^2 = 0.04 \times 64 \times 10^{12} = 2.56 \times 10^{12}$
$v = \sqrt{2.56 \times 10^{12}} = 1.6 \times 10^6 \ m \ s^{-1} = 1600 \ km \ s^{-1}$.

(D) Given: mass $m = 1 \text{ g} = 10^{-3} \text{ kg}$, charge $q = 20 \mu\text{C} = 20 \times 10^{-6} \text{ C}$, length $r = 0.9 \text{ m}$, electric field $E = 100 \text{ NC}^{-1}$, and $g = 10 \text{ ms}^{-2}$.
The force exerted by the electric field on the charge is $F_e = qE = 20 \times 10^{-6} \times 100 = 2 \times 10^{-3} \text{ N}$ (upwards).
The gravitational force on the ball is $F_g = mg = 10^{-3} \times 10 = 10 \times 10^{-3} \text{ N}$ (downwards).
The net effective force acting downwards is $F_{\text{eff}} = F_g - F_e = 10 \times 10^{-3} - 2 \times 10^{-3} = 8 \times 10^{-3} \text{ N}$.
The effective acceleration due to gravity is $g_{\text{eff}} = \frac{F_{\text{eff}}}{m} = \frac{8 \times 10^{-3}}{10^{-3}} = 8 \text{ ms}^{-2}$.
For a particle to complete a vertical circle, the minimum velocity at the lowest point is $v = \sqrt{5g_{\text{eff}}r}$.
Substituting the values: $v = \sqrt{5 \times 8 \times 0.9} = \sqrt{40 \times 0.9} = \sqrt{36} = 6 \text{ ms}^{-1}$.

(D) Let $v$ be the initial velocity. The change in momentum is $\Delta \vec{p} = q \vec{E} \Delta t$. Since $\vec{E}$ is uniform,$\Delta \vec{p}$ is the same for both balls if we assume the same $qE\Delta t$ or relate them via ratios. For the first ball,the final velocity $\vec{v}_1$ has magnitude $v/2$ and is at $60^{\circ}$ to $\vec{v}$. The change in momentum $\Delta \vec{p}_1 = m_1(\vec{v}_1 - \vec{v})$ must be perpendicular to $\vec{v}_1$ because the magnitude decreased. Using the law of cosines: $(v/2)^2 = v^2 + (\Delta p_1/m_1)^2 - 2v(\Delta p_1/m_1)\cos(120^{\circ})$. Solving this,$\Delta p_1 = m_1 v \sin(60^{\circ}) = m_1 v \frac{\sqrt{3}}{2}$. For the second ball,the velocity changes by $90^{\circ}$,so $\vec{v}_2 \perp \vec{v}$. Thus,$v_2 = v / \cos(90^{\circ})$ is not applicable; rather,$\Delta p_2 = m_2 v \tan(90^{\circ})$ implies the force is perpendicular to the initial velocity. The magnitude of the second ball's velocity is $v_2 = v / \cos(30^{\circ}) = 2v/\sqrt{3}$ is incorrect; the correct relation is $v_2 = v \tan(30^{\circ}) = v/\sqrt{3}$. The ratio $\frac{q_2/m_2}{q_1/m_1} = \frac{4}{3}$,so $k_2 = \frac{4}{3} k_1$.

(D) The acceleration $a$ of a particle of mass $m$ and charge $q$ in a uniform electric field $E$ is given by $a = \frac{qE}{m}$.
For a proton $(p)$: $q_p = e$, $m_p = m$. Thus, $a_p = \frac{eE}{m}$.
For an $\alpha$-particle $(\alpha)$: $q_{\alpha} = 2e$, $m_{\alpha} = 4m$. Thus, $a_{\alpha} = \frac{2eE}{4m} = \frac{eE}{2m}$.
Since they start from rest, the distance $s$ traveled in time $t$ is $s = \frac{1}{2}at^2$. Given $s$ is the same for both, $\frac{1}{2}a_p t_p^2 = \frac{1}{2}a_{\alpha} t_{\alpha}^2$.
$\frac{t_p^2}{t_{\alpha}^2} = \frac{a_{\alpha}}{a_p} = \frac{eE/2m}{eE/m} = \frac{1}{2}$.
Therefore, the ratio of times taken by the proton to the $\alpha$-particle is $\frac{t_p}{t_{\alpha}} = \frac{1}{\sqrt{2}}$.

(D) Let the masses be $m_1 = m$ and $m_2 = 3m$. The ratio of masses is $m_1 : m_2 = 1 : 3$.
The charges are in reciprocal ratio to their masses,so $q_1 : q_2 = 3 : 1$. Let $q_1 = 3q$ and $q_2 = q$.
When placed in a uniform electric field $E$,the force on each particle is $F = qE$.
The acceleration of each particle is $a = F/m = qE/m$.
Assuming they start from rest,after time $t$,the velocity is $v = at = (qE/m)t$.
The kinetic energy is $K = (1/2)mv^2 = (1/2)m(qEt/m)^2 = (q^2 E^2 t^2) / (2m)$.
The ratio of kinetic energies is $K_1 / K_2 = [(q_1^2) / (2m_1)] / [(q_2^2) / (2m_2)] = (q_1/q_2)^2 * (m_2/m_1)$.
Substituting the values: $K_1 / K_2 = (3/1)^2 * (3/1) = 9 * 3 = 27 / 1$.
Thus,the ratio is $27: 1$.

(B) Initially,the bead is at rest,which implies that the upward electric force $F_e = qE$ balances the downward gravitational force $mg$. Thus,$qE = mg$.
When the electric field is switched off,the only force acting on the bead is gravity $(mg)$. The bead will start moving downwards with a constant acceleration $g$.
When the electric field is switched on again,the electric force $F_e = qE$ acts upwards again. Since the bead has gained some downward velocity during the time the field was off,it will continue to move downwards while decelerating until its velocity becomes zero.
After the velocity becomes zero,the electric force $F_e$ (which is equal to $mg$) will cause the bead to accelerate upwards until it returns to its original position.
Therefore,the bead moves downwards and then moves upwards.

(D) The electric field $E$ acts upwards. Since the charge $q$ is negative,the force $F = qE$ acts downwards. The acceleration of the particle is $a = \frac{|q|E}{m} = \frac{1 \times 10^{-6} \times 10^3}{2 \times 10^{-3}} = 0.5 \ ms^{-2}$ downwards.
For the particle not to hit the upper plate,its maximum vertical displacement $h_{\max}$ must be less than or equal to the plate separation $d = 2 \ cm = 0.02 \ m$.
The vertical component of velocity is $u_y = u \sin 45^{\circ} = \frac{u}{\sqrt{2}}$.
At maximum height,the vertical velocity becomes zero. Using $v_y^2 = u_y^2 - 2ah_{\max}$,we get $0 = (\frac{u}{\sqrt{2}})^2 - 2ah_{\max}$,so $h_{\max} = \frac{u^2}{4a}$.
Setting $h_{\max} = 0.02 \ m$ and $a = 0.5 \ ms^{-2}$:
$0.02 = \frac{u^2}{4 \times 0.5} = \frac{u^2}{2}$
$u^2 = 0.04 \implies u = 0.2 \ ms^{-1}$.

(B) The force experienced by an electron of charge $e$ in an electric field $E$ is given by $F = eE$.
According to Newton's second law,the acceleration $a$ of the electron is $a = \frac{F}{m_e} = \frac{eE}{m_e}$.
Since the electron is released from rest,its initial velocity $u = 0$.
Using the kinematic equation for distance $S$ travelled in time $t$:
$S = ut + \frac{1}{2}at^2$
$S = 0 \cdot t + \frac{1}{2} \left( \frac{eE}{m_e} \right) t^2$
$S = \frac{eEt^2}{2m_e}$.

(D) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K.E. = eV$.
Since the electron starts from rest,the kinetic energy is equal to $\frac{1}{2}mv^2$.
Equating the two: $\frac{1}{2}mv^2 = eV$.
Solving for velocity $v$: $v = \sqrt{\frac{2eV}{m}}$.
Given: $e = 1.6 \times 10^{-19} \ C$,$V = 180 \ V$,$m = 9 \times 10^{-31} \ kg$.
Substituting the values: $v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 180}{9 \times 10^{-31}}}$.
$v = \sqrt{\frac{576 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{64 \times 10^{12}} = 8 \times 10^6 \ m/s$.
Converting to $km/s$: $v = 8000 \ km/s$.

(A) Given: Mass $m = 2 \ g = 2 \times 10^{-3} \ kg$,Charge $q = 6 \ \mu C = 6 \times 10^{-6} \ C$,Potential difference $V = 60 \ V$.
According to the work-energy theorem,the kinetic energy gained by the particle is equal to the work done by the electric field:
$K.E. = qV$
$\frac{1}{2}mv^2 = qV$
$v^2 = \frac{2qV}{m}$
$v = \sqrt{\frac{2qV}{m}}$
Substituting the values:
$v = \sqrt{\frac{2 \times (6 \times 10^{-6} \ C) \times (60 \ V)}{2 \times 10^{-3} \ kg}}$
$v = \sqrt{\frac{720 \times 10^{-6}}{2 \times 10^{-3}}}$
$v = \sqrt{360 \times 10^{-3}} = \sqrt{0.36} = 0.6 \ ms^{-1}$.

(B) The electric field is uniform and directed along the positive $x$-axis,so $\vec{E} = E \hat{i}$.
The electric force on the charge $q$ is $\vec{F} = q \vec{E} = q E \hat{i}$.
Since the electric force is a conservative force,the work done depends only on the initial and final positions,not on the path taken.
The initial position is $P(a, b, 0)$ and the final position is $S(0, 0, 0)$.
The displacement vector is $\vec{d} = \vec{S} - \vec{P} = (0 - a) \hat{i} + (0 - b) \hat{j} + (0 - 0) \hat{k} = -a \hat{i} - b \hat{j}$.
The work done by the electric field is $W = \vec{F} \cdot \vec{d}$.
$W = (q E \hat{i}) \cdot (-a \hat{i} - b \hat{j}) = -q E a (\hat{i} \cdot \hat{i}) - q E b (\hat{i} \cdot \hat{j})$.
Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = 0$,we get $W = -q E a$.

(D) Let the length of the string be $L$.
Using the work-energy theorem,the work done by the electric force is equal to the change in kinetic energy:
$W = \Delta K
\Rightarrow qEL = \frac{1}{2}mv^2 - 0
\Rightarrow v^2 = \frac{2qEL}{m}$.
When the string becomes parallel to the electric field,the forces acting on the block are the tension $T$ (inward) and the electric force $qE$ (outward).
The net centripetal force is given by:
$T - qE = \frac{mv^2}{L}$.
Substituting the value of $v^2$:
$T = qE + \frac{m}{L} \left( \frac{2qEL}{m} \right)
\Rightarrow T = qE + 2qE
\Rightarrow T = 3qE$.

(D) Initially,the oil drop is in equilibrium,so the electric force balances the gravitational force: $qE = mg$.
Here,$m = 4.8 \times 10^{-10} \,g = 4.8 \times 10^{-13} \,kg$,$q = 2.4 \times 10^{-18} \,C$,and $g = 10 \,m/s^2$.
When the polarity of the plates is changed,the direction of the electric force $qE$ reverses and acts downwards along with the gravitational force $mg$.
The new net force on the drop is $F_{net} = qE + mg$.
Since $qE = mg$,we have $F_{net} = mg + mg = 2mg$.
The instantaneous acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{2mg}{m} = 2g$.
Substituting $g = 10 \,m/s^2$,we get $a = 2 \times 10 = 20 \,m/s^2$.

(D) The force acting on the particle in the electric field is $F = qE$.
According to Newton's second law,the acceleration of the particle in the direction of the field is $a = \frac{F}{m} = \frac{qE}{m}$.
Since the particle is thrown perpendicular to the field,the initial velocity component in the direction of the field is $0$.
Along the $x$-direction (perpendicular to the field),there is no acceleration,so the velocity remains constant at $v$. Thus,the distance covered in time $t$ is $x = vt$,which gives $t = \frac{x}{v}$.
Along the $y$-direction (parallel to the field),using the second equation of motion $y = u_y t + \frac{1}{2} a_y t^2$,where $u_y = 0$ and $a_y = a = \frac{qE}{m}$,we get:
$y = 0 + \frac{1}{2} \left( \frac{qE}{m} \right) \left( \frac{x}{v} \right)^2$.
$y = \frac{qE}{2mv^2} x^2$.
Comparing this with the given equation $y = \alpha x^2$,we find $\alpha = \frac{qE}{2mv^2}$.

(A) The deflection $y$ of a charged particle moving with velocity $u$ in a uniform electric field $E$ over a distance $L$ is given by $y = \frac{1}{2} a t^2 = \frac{1}{2} (\frac{qE}{m}) (\frac{L}{u})^2 = \frac{qEL^2}{2mu^2}$.
Since momentum $p = mu$, we have $u = p/m$. Substituting this, $y = \frac{qEL^2}{2m(p/m)^2} = \frac{qEL^2m}{2p^2}$.
Given that $E, L,$ and $p$ are constant for all particles, the deflection $y \propto qm$.
For a proton $(p)$, deuteron $(d)$, and $\alpha$-particle $(\alpha)$:
Charge ratio: $q_p : q_d : q_\alpha = 1 : 1 : 2$.
Mass ratio: $m_p : m_d : m_\alpha = 1 : 2 : 4$.
Therefore, the ratio of deflections is $y_p : y_d : y_\alpha = (q_p m_p) : (q_d m_d) : (q_\alpha m_\alpha) = (1 \times 1) : (1 \times 2) : (2 \times 4) = 1 : 2 : 8$.

(B) The force on a charge $q$ in an electric field $E$ is given by $F = qE$.
For an electron,$q = -e$,so the force is $F_e = -eE$. The acceleration is $a_e = \frac{-eE}{m}$.
For a positron,$q = +e$,so the force is $F_p = +eE$. The acceleration is $a_p = \frac{eE}{m}$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,and since the initial velocity in the direction of the field is zero $(u = 0)$:
Displacement of the electron in the field direction: $y_e = \frac{1}{2} a_e t^2 = -\frac{eE t^2}{2m}$.
Displacement of the positron in the field direction: $y_p = \frac{1}{2} a_p t^2 = \frac{eE t^2}{2m}$.
The separation distance between them in the direction of the field is $d = |y_p - y_e| = |\frac{eE t^2}{2m} - (-\frac{eE t^2}{2m})| = \frac{eE t^2}{m}$.

(D) Given: Mass $m = 10 \ mg = 10 \times 10^{-6} \ kg = 10^{-5} \ kg$,Charge $q = 2 \ \mu C = 2 \times 10^{-6} \ C$,Potential difference $V = 160 \ V$.
According to the work-energy theorem,the work done by the electric field is equal to the change in kinetic energy of the particle.
$W = \Delta K$
$qV = \frac{1}{2}mv^2 - 0$
$v^2 = \frac{2qV}{m}$
$v^2 = \frac{2 \times (2 \times 10^{-6} \ C) \times 160 \ V}{10^{-5} \ kg}$
$v^2 = \frac{640 \times 10^{-6}}{10^{-5}} = 640 \times 10^{-1} = 64$
$v = \sqrt{64} = 8 \ ms^{-1}$.

(C) The acceleration $a$ of a particle of charge $q$ and mass $m$ in a uniform electric field $E$ is given by $a = \frac{qE}{m}$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,since the particles start from rest $(u = 0)$,we have $s = \frac{1}{2}at^2$.
Thus,$t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2sm}{qE}}$.
For equal displacements $s$,the time $t$ is proportional to $\sqrt{\frac{m}{q}}$.
Let $m_p$ and $q_p$ be the mass and charge of the proton,and $m_\alpha$ and $q_\alpha$ be the mass and charge of the alpha particle.
We know $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$.
The ratio of times is $\frac{t_p}{t_\alpha} = \sqrt{\frac{m_p}{q_p} \cdot \frac{q_\alpha}{m_\alpha}} = \sqrt{\frac{m_p}{q_p} \cdot \frac{2q_p}{4m_p}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

(C) For the block to remain at rest on a frictionless inclined plane, the component of gravitational force acting down the plane must be balanced by the electric force acting up the plane.
$mg \sin \theta = qE$
$E = \frac{mg \sin \theta}{q}$
Given: $m = 5 \, g = 5 \times 10^{-3} \, kg$, $q = 5 \, \mu C = 5 \times 10^{-6} \, C$, $\theta = 60^{\circ}$, $g = 10 \, m/s^2$.
$E = \frac{5 \times 10^{-3} \times 10 \times \sin(60^{\circ})}{5 \times 10^{-6}}$
$E = \frac{5 \times 10^{-2} \times \frac{\sqrt{3}}{2}}{5 \times 10^{-6}}$
$E = 10^4 \times \frac{\sqrt{3}}{2} \, N/C$
Thus, the magnitude of the electric field is $\frac{\sqrt{3}}{2} \times 10^4 \, N/C$.

(B) proton is a positively charged particle. When placed in an electric field $E$,it experiences an electrostatic force $F = qE$ in the direction of the electric field.
To move the proton against this Coulomb force,an external force must be applied to perform work on the proton.
Since the electric field naturally pushes the proton in its own direction,moving it in the opposite direction requires an external agency to supply energy to the system.
Therefore,energy is used from some outside source to perform this work.

(C) Given: Electric field $E = 10^3 \ Vm^{-1}$ along $Y$-axis,mass $m = 1 \ g = 10^{-3} \ kg$,charge $q = 10^{-6} \ C$,initial velocity $u_x = 10 \ ms^{-1}$ along $X$-axis.
The force on the charge is $F = qE = 10^{-6} \times 10^3 = 10^{-3} \ N$ along the $Y$-axis.
The acceleration along the $Y$-axis is $a_y = F/m = 10^{-3} / 10^{-3} = 1 \ ms^{-2}$.
The velocity along the $X$-axis remains constant as there is no force in that direction: $v_x = u_x = 10 \ ms^{-1}$.
The velocity along the $Y$-axis after $t = 10 \ s$ is $v_y = u_y + a_y t = 0 + 1 \times 10 = 10 \ ms^{-1}$.
The final speed is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10 \sqrt{2} \ ms^{-1}$.

(C) The particle moves with constant velocity $v$ along the $x$-axis,so $x = vt$,which implies $t = x/v$.
Along the $y$-axis,the particle experiences a constant force $F = qE$,leading to an acceleration $a = qE/m$.
Using the kinematic equation $y = u_y t + \frac{1}{2} a_y t^2$,where $u_y = 0$:
$y = \frac{1}{2} (\frac{qE}{m}) (\frac{x}{v})^2 = (\frac{qE}{2mv^2}) x^2$.
This equation is of the form $y = kx^2$,which represents a parabola.
Therefore,the trajectory of the particle is parabolic.

(A) The acceleration of a charged particle in a uniform electric field $E$ is given by $a = \frac{qE}{m}$.
Since the particles start from rest $(u = 0)$,the velocity after time $t$ is $v = at = \left(\frac{qE}{m}\right)t$.
According to the work-energy theorem,the work done $W$ by the electric field is equal to the change in kinetic energy $\Delta K$:
$W = \Delta K = \frac{1}{2}mv^2$.
Substituting the expression for $v$:
$W = \frac{1}{2}m\left(\frac{qEt}{m}\right)^2 = \frac{q^2 E^2 t^2}{2m}$.
Since $E$ and $t$ are the same for both particles,$W \propto \frac{q^2}{m}$.
For a proton,$q_p = e$ and $m_p = m$. For an $\alpha$-particle,$q_\alpha = 2e$ and $m_\alpha = 4m$.
Therefore,the ratio is:
$\frac{W_p}{W_\alpha} = \left(\frac{q_p}{q_\alpha}\right)^2 \left(\frac{m_\alpha}{m_p}\right) = \left(\frac{e}{2e}\right)^2 \left(\frac{4m}{m}\right) = \left(\frac{1}{4}\right) \times 4 = 1: 1$.

(B) The force on the particle is $F = qE = qE_0 \sin(\omega t)$.
Using Newton's second law,$ma = qE_0 \sin(\omega t)$,so $a = \frac{qE_0}{m} \sin(\omega t)$.
The velocity $v(t)$ is the integral of acceleration: $v(t) = \int a \, dt = \int \frac{qE_0}{m} \sin(\omega t) \, dt = -\frac{qE_0}{m\omega} \cos(\omega t) + C$.
Since the particle starts from rest at $t = 0$,$v(0) = 0$,which gives $C = \frac{qE_0}{m\omega}$.
Thus,$v(t) = \frac{qE_0}{m\omega} (1 - \cos(\omega t))$.
The maximum speed occurs when $\cos(\omega t) = -1$,so $v_{\max} = \frac{2qE_0}{m\omega}$.
Substituting the values: $q = 1 \text{ C}$,$E_0 = 2 \text{ N/C}$,$m = 1 \text{ g} = 10^{-3} \text{ kg}$,and $\omega = 1000 \text{ rad/s}$.
$v_{\max} = \frac{2 \times 1 \times 2}{10^{-3} \times 1000} = \frac{4}{1} = 4 \text{ m/s}$.

(B) The electric field between the plates is $E = \frac{V}{d}$.
The force on an electron of charge $e$ is $F = eE = \frac{eV}{d}$.
The acceleration of the electron in the transverse direction ($y$-axis) is $a_y = \frac{F}{m} = \frac{eV}{md}$.
The time taken to travel an axial distance $L$ with constant horizontal velocity $v_0$ is $t = \frac{L}{v_0}$.
The transverse velocity $v_y$ acquired after time $t$ is $v_y = a_y t = \left(\frac{eV}{md}\right) \left(\frac{L}{v_0}\right) = \frac{eVL}{mdv_0}$.
The horizontal velocity remains constant at $v_x = v_0$.
The angle $\theta$ that the beam makes with the plates is given by $\tan \theta = \frac{v_y}{v_x}$.
Substituting the values,$\tan \theta = \frac{eVL / mdv_0}{v_0} = \frac{eVL}{mdv_0^2}$.
Therefore,$\theta = \tan ^{-1}\left(\frac{eVL}{mdv_0^2}\right)$.