Motion of Charge particle in Electric filed Questions in English

(C) The force experienced by a charged particle in a uniform electric field $E$ is given by $F = qE$.
According to Newton's second law,$F = ma$,so the acceleration is $a = \frac{qE}{m}$.
Since both the electron and the proton have the same magnitude of charge $e$,the acceleration is inversely proportional to the mass $(a \propto \frac{1}{m})$.
Therefore,the ratio of the acceleration of the electron $(a_e)$ to the acceleration of the proton $(a_p)$ is $\frac{a_e}{a_p} = \frac{m_p}{m_e}$.
Thus,the ratio of their accelerations is the ratio of the mass of the proton to the mass of the electron.

(B) The electric field $E$ at any point is directed along the tangent to the electric field line at that point.
If a charged particle is placed in an electric field with an initial velocity of $v = 0$,the force $F = qE$ acts on it in the direction of the electric field.
Since the force is tangent to the field line,the particle will accelerate along the field line.
If the particle has an initial velocity at an angle to the field line,the trajectory will generally be curved and not follow the field line.
Therefore,the particle travels along a line of force only if its initial velocity is zero.

(A) To balance the proton,the upward electric force must equal the downward gravitational force.
$qE = mg$
$E = \frac{mg}{q}$
Given:
$m = 1.7 \times 10^{-27} \ kg$
$q = 1.6 \times 10^{-19} \ C$
$g = 9.8 \ m/s^2$
Substituting the values:
$E = \frac{1.7 \times 10^{-27} \times 9.8}{1.6 \times 10^{-19}}$
$E \approx \frac{1.666 \times 10^{-26}}{1.6 \times 10^{-19}}$
$E \approx 1.04 \times 10^{-7} \ V/m$
Thus,the intensity is nearly $1 \times 10^{-7} \ V/m$.

(B) Given: Mass $m = 20\,g = 20 \times 10^{-3}\,kg$,Charge $q = 3.0\,mC = 3.0 \times 10^{-3}\,C$,Initial velocity $u = 20\,m/s$,Electric field $E = 80\,N/C$,Time $t = 3\,s$.
The force acting on the charge is $F = qE$.
According to Newton's second law,$F = ma$,so $a = \frac{qE}{m}$.
Substituting the values: $a = \frac{3.0 \times 10^{-3} \times 80}{20 \times 10^{-3}} = \frac{240 \times 10^{-3}}{20 \times 10^{-3}} = 12\,m/s^2$.
Using the first equation of motion: $v = u + at$.
$v = 20 + (12 \times 3) = 20 + 36 = 56\,m/s$.

(B) For the particle to be in equilibrium,the upward electric force must balance the downward gravitational force.
$QE = Mg$
Since $E = \frac{V}{d}$,we have $Q \left( \frac{V}{d} \right) = Mg$.
Given $Q = Ne$,where $N$ is the number of elementary charges:
$Ne \left( \frac{V}{d} \right) = Mg$
$N = \frac{Mgd}{eV}$
Substituting the given values ($M = 1.96 \times 10^{-15} \; kg$,$g = 9.8 \; m/s^2$,$d = 0.02 \; m$,$V = 800 \; V$,$e = 1.6 \times 10^{-19} \; C$):
$N = \frac{1.96 \times 10^{-15} \times 9.8 \times 0.02}{1.6 \times 10^{-19} \times 800}$
$N = \frac{3.8416 \times 10^{-16}}{1.28 \times 10^{-16}} = 3$
Therefore,the charge on the particle is $Q = 3e$.

(C) The force experienced by a charge $q$ in a uniform electric field $E$ is given by $F = qE$.
Since the particle starts from rest and moves a distance $y$ in the direction of the force,the work done by the electric field on the particle is $W = F \times y = (qE) \times y = qEy$.
According to the work-energy theorem,the work done by the electric field is equal to the change in kinetic energy of the particle.
Since the initial kinetic energy is $0$,the final kinetic energy attained by the particle is $K = qEy$.

(B) The force on a charged particle in a uniform electric field $E$ is given by $F = qE$. According to Newton's second law,$F = ma$,so the acceleration is $a = qE/m$.
Using the equation of motion $s = ut + (1/2)at^2$,and since the particles start from rest $(u = 0)$,the distance $s$ covered in time $t$ is $s = (1/2)(qE/m)t^2$.
For the electron: $s = (1/2)(eE/m_e)t_1^2$.
For the proton: $s = (1/2)(eE/m_p)t_2^2$.
Since the distance $s$ is the same for both,we equate the two expressions:
$(1/2)(eE/m_e)t_1^2 = (1/2)(eE/m_p)t_2^2$.
Simplifying this,we get $t_1^2/m_e = t_2^2/m_p$.
Therefore,$t_2^2/t_1^2 = m_p/m_e$.
Taking the square root of both sides,we find the ratio $t_2/t_1 = (m_p/m_e)^{1/2}$.

(C) For the water drop to be in equilibrium,the electric force must balance the gravitational force.
$QE = mg$
Here,$Q = e = 1.6 \times 10^{-19}\,C$,$r = 0.1\,\mu m = 10^{-7}\,m$,$\rho = 10^3\,kg/m^3$ (density of water),and $g = 10\,m/s^2$.
The mass $m$ is given by $m = \text{Volume} \times \text{Density} = \frac{4}{3}\pi r^3 \rho$.
Substituting the values:
$E = \frac{mg}{Q} = \frac{\frac{4}{3} \times 3.14 \times (10^{-7})^3 \times 10^3 \times 10}{1.6 \times 10^{-19}}$
$E = \frac{4 \times 3.14 \times 10^{-21} \times 10^4}{3 \times 1.6 \times 10^{-19}}$
$E = \frac{12.56 \times 10^{-17}}{4.8 \times 10^{-19}} = 2.616 \times 10^2 \approx 262\,N/C$.

(B) For the oil drop to be stationary,the electric force must balance the gravitational force.
$F_e = F_g$
$qE = mg$
Since $E = \frac{V}{d}$ and $m = \text{Volume} \times \text{Density} = \frac{4}{3} \pi r^3 \rho$,we have:
$q \left( \frac{V}{d} \right) = \left( \frac{4}{3} \pi r^3 \rho \right) g$
Given: $q = 2e = 2 \times 1.6 \times 10^{-19} \, C$,$V = 12000 \, V$,$d = 2.0 \times 10^{-2} \, m$,$\rho = 900 \, kg/m^3$,$g = 10 \, m/s^2$.
Substituting the values:
$2 \times 1.6 \times 10^{-19} \times \frac{12000}{2 \times 10^{-2}} = \frac{4}{3} \times 3.14 \times r^3 \times 900 \times 10$
$1.92 \times 10^{-13} = 37680 \times r^3$
$r^3 = \frac{1.92 \times 10^{-13}}{37680} \approx 0.509 \times 10^{-17} = 5.09 \times 10^{-18} \, m^3$
$r = \sqrt[3]{5.09 \times 10^{-18}} \approx 1.72 \times 10^{-6} \, m$.

(C) The kinetic energy gained by a charged particle accelerated through a potential difference $V$ is given by the formula $\Delta KE = qV$.
For a proton,the charge $q = e$.
Given the potential difference $V = 1\,V$.
Substituting these values,we get $\Delta KE = e \times 1\,V = 1\,eV$.
Therefore,the kinetic energy of the proton is $1\,eV$.

(B) The force $F$ on an electron in an electric field $E$ is given by $F = qE$.
The electric field between two parallel plates is $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the separation distance.
Given: $q = 1.6 \times 10^{-19}\,C$,$V = 1000\,V$,and $d = 2\,mm = 2 \times 10^{-3}\,m$.
Substituting these values into the formula:
$F = q \left( \frac{V}{d} \right) = (1.6 \times 10^{-19}) \left( \frac{1000}{2 \times 10^{-3}} \right)$
$F = (1.6 \times 10^{-19}) \times (0.5 \times 10^{6})$
$F = 0.8 \times 10^{-13}\,N = 8 \times 10^{-14}\,N$.

(B) For the particle to be held stationary,the upward electric force must balance the downward gravitational force.
$F_e = F_g$
$|q|E = mg$
$|q| = \frac{mg}{E}$
Given: $m = 5 \times 10^{-5} \ kg$,$g = 10 \ m/s^2$,$E = 10^7 \ N/C$.
$|q| = \frac{5 \times 10^{-5} \times 10}{10^7} = 5 \times 10^{-11} \ C$.
Since $1 \ C = 10^6 \ \mu C$,we have $|q| = 5 \times 10^{-11} \times 10^6 \ \mu C = 5 \times 10^{-5} \ \mu C$.
Since the electric field is directed downwards and the gravitational force acts downwards,the electric force must act upwards to balance it. For a downward electric field,a negative charge experiences an upward force. Therefore,the charge is $-5 \times 10^{-5} \ \mu C$.

(C) When a charged particle like an electron enters a uniform electric field perpendicularly to the direction of the field,it experiences a constant force in the direction of the field (or opposite to it).
Since the velocity along the $x$-axis remains constant and the acceleration along the $y$-axis is constant,the equation of motion follows the form $y = kx^2$.
This is the equation of a parabola. Therefore,the path of the electron is parabolic.

(C) The electric field $E$ exerts a force $F = qE$ on a charged particle. For an electron,the charge $q = -e$,so the force $F = -eE$ acts in the direction opposite to the electric field lines.
Since the electron enters the field with its velocity in the direction of the electric lines of force,the force acting on it is directed opposite to its velocity.
This force acts as a retarding force,causing the electron to decelerate.
Therefore,the velocity of the electron will decrease.

(C) When an electron of charge $e$ is accelerated through a potential difference $V$,the work done on the electron is $W = eV$.
This work done is converted into the kinetic energy of the electron.
According to the work-energy theorem,$K = \frac{1}{2}mv^2 = eV$.
Solving for the final speed $v$,we get $v^2 = \frac{2eV}{m}$.
Therefore,the final speed is $v = \sqrt{\frac{2eV}{m}}$.

(A) The force on an electron in an electric field is given by $F = eE$.
According to Newton's second law,$F = ma$,so $ma = eE$.
Therefore,the acceleration $a$ is given by $a = (e/m) \times E$.
Given:
$e/m = 1.76 \times 10^{11}\, C/kg$
$E = 50\, V/cm = 50 \times 10^2\, V/m = 5000\, V/m$.
Substituting the values:
$a = (1.76 \times 10^{11}) \times (5000) = 1.76 \times 10^{11} \times 5 \times 10^3 = 8.8 \times 10^{14}\, m/s^2$.

(A) When a charged particle enters a uniform electric field $\vec{E}$ directed along the $y$-axis,it experiences an electric force $\vec{F} = q\vec{E}$ in the direction of the $y$-axis.
According to Newton's second law,this force causes an acceleration $a_y = \frac{qE}{m}$ in the $y$-direction,which changes the vertical component of the velocity.
Since there is no electric field component along the $x$-axis,there is no force acting in the $x$-direction $(F_x = 0)$.
Therefore,the horizontal component of the velocity remains constant throughout the motion.
Thus,the vertical velocity changes while the horizontal velocity remains constant.

(A) The kinetic energy $(KE)$ gained by a charged particle accelerated through a potential difference $(V)$ is given by the formula: $KE = qV$.
Here,the charge of an electron is $q = 1.6 \times 10^{-19} \, C$ and the potential difference is $V = 100 \, V$.
Substituting these values into the formula:
$KE = (1.6 \times 10^{-19} \, C) \times (100 \, V)$
$KE = 1.6 \times 10^{-19} \times 10^2 \, J$
$KE = 1.6 \times 10^{-17} \, J$.
Therefore,the correct option is $A$.

(A) To balance the weight of the water drop,the electric force must act in the upward direction to counteract the gravitational force acting downward.
Given:
Mass $m = 10^{-6} \ kg$
Charge $q = 10^{-6} \ C$
Acceleration due to gravity $g = 10 \ m/s^2$
The condition for equilibrium is:
$F_e = W$
$qE = mg$
$E = \frac{mg}{q}$
Substituting the values:
$E = \frac{10^{-6} \times 10}{10^{-6}} = 10 \ V/m$
Since the charge is positive,the electric force is in the direction of the electric field. To balance the downward weight,the electric field must be applied in the upward direction.

(B) For the particle to be held stationary,the upward electric force must balance the downward gravitational force.
$F_e = F_g$
$QE = mg$
Given:
Mass $m = 0.003 \, g = 0.003 \times 10^{-3} \, kg = 3 \times 10^{-6} \, kg$
Electric field $E = 6 \times 10^4 \, N/C$
Acceleration due to gravity $g = 10 \, m/s^2$
$Q = \frac{mg}{E} = \frac{3 \times 10^{-6} \times 10}{6 \times 10^4}$
$Q = \frac{3 \times 10^{-5}}{6 \times 10^4} = 0.5 \times 10^{-9} \, C = 5 \times 10^{-10} \, C$

(C) The force $F$ experienced by a charge $q$ in an electric field $E$ is given by $F = qE$.
For an electron,the charge is $q = e$,so the force is $F = eE$.
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: $ma = eE$.
Therefore,the acceleration $a$ is given by $a = \frac{eE}{m}$.

(C) The kinetic energy $(KE)$ gained by a charged particle accelerated through a potential difference $(V)$ is given by the formula $KE = qV$.
For an $\alpha$-particle,the charge is $q = +2e$,where $e$ is the elementary charge.
Given the potential difference $V = 200\,V$.
Substituting these values into the formula:
$KE = (2e) \times 200\,V = 400\,eV$.
Therefore,the increase in its kinetic energy is $400\,eV$.

(C) The electric force acting on the electron is $F = qE = ma$.
Thus,the retardation $a = \frac{qE}{m}$.
Substituting the given values:
$a = \frac{1.6 \times 10^{-19} \times 1 \times 10^3}{9 \times 10^{-31}} = \frac{1.6}{9} \times 10^{15} \, m/s^2$.
Given initial velocity $u = 5 \times 10^6 \, m/s$ and final velocity $v = 0$.
Using the kinematic equation $v^2 = u^2 - 2as$,we find the distance $s = \frac{u^2}{2a}$.
$s = \frac{(5 \times 10^6)^2}{2 \times (\frac{1.6}{9} \times 10^{15})} = \frac{25 \times 10^{12} \times 9}{3.2 \times 10^{15}} = \frac{225}{3.2} \times 10^{-3} \approx 70.3 \times 10^{-3} \, m = 7.03 \times 10^{-2} \, m \approx 7 \, cm$.

(A) For a charged particle to be in equilibrium in a vertical electric field,the upward electric force must balance the downward gravitational force.
$F_e = F_g$
$qE = mg$
$q = \frac{mg}{E} = \frac{9.6 \times 10^{-16} \times 10}{20,000} = \frac{9.6 \times 10^{-15}}{2 \times 10^4} = 4.8 \times 10^{-19}\, C$.
Since $q = ne$,the number of excess electrons $n$ is:
$n = \frac{q}{e} = \frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}} = 3$.
Thus,the charge is $4.8 \times 10^{-19}\, C$ and the number of electrons is $3$.

(C) The force experienced by an electron in an electric field is given by $F = qE$,where $q = e$ is the charge of the electron and $E$ is the electric field intensity.
Given that the force is equal to the weight of the electron,we have $F = mg$,where $m$ is the mass of the electron $(m \approx 9.1 \times 10^{-31} \ kg)$ and $g$ is the acceleration due to gravity $(g \approx 9.8 \ m/s^2)$.
Equating the two,we get $eE = mg$,which implies $E = \frac{mg}{e}$.
Substituting the values: $E = \frac{9.1 \times 10^{-31} \ kg \times 9.8 \ m/s^2}{1.6 \times 10^{-19} \ C}$.
$E \approx \frac{89.18 \times 10^{-31}}{1.6 \times 10^{-19}} \ N/C \approx 5.57 \times 10^{-11} \ N/C$.
Rounding to the nearest given option,the intensity of the field is $5.5 \times 10^{-11} \ N/C$.

(C) The initial time period of the pendulum is $T_0 = 2\pi \sqrt{\frac{L}{g}}$.
When the plates are charged,an electric field $E$ is created. The force on the charge $q$ is $F_e = qE$ acting downwards.
The net downward force on the sphere is $F_{net} = mg + qE$.
The effective acceleration $g'$ is given by $g' = \frac{F_{net}}{m} = g + \frac{qE}{m}$.
The new time period is $T = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}$.
Taking the ratio $T/T_0$,we get:
$\frac{T}{T_0} = \frac{2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{g + \frac{qE}{m}}} = \left( \frac{g}{g + \frac{qE}{m}} \right)^{1/2}$.
Thus,the correct option is $(c)$.

(D) By the law of conservation of energy,the initial kinetic energy of the particle $q$ is converted into electrostatic potential energy at the point of closest approach,where the particle momentarily comes to rest.
Initial kinetic energy = $\frac{1}{2}mv^2$
Electrostatic potential energy at distance $r$ = $\frac{1}{4\pi\varepsilon_0} \cdot \frac{qQ}{r}$
Equating the two: $\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{qQ}{r}$
From this,we can see that $r = \frac{qQ}{2\pi\varepsilon_0 mv^2}$,which implies $r \propto \frac{1}{v^2}$.
If the speed is doubled $(v' = 2v)$,the new distance of closest approach $r'$ will be:
$r' = r \cdot \left(\frac{v}{v'}\right)^2 = r \cdot \left(\frac{v}{2v}\right)^2 = r \cdot \frac{1}{4} = \frac{r}{4}$.
Thus,the closest distance of approach becomes $r/4$.

(A) The work done by an electric field on a moving charge is given by $W = qEd$,which results in a change in kinetic energy.
In contrast,the magnetic force acting on a moving charge is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$.
Since the magnetic force is always perpendicular to the velocity of the charge,the power delivered by the magnetic field is $P = \vec{F}_m \cdot \vec{v} = 0$.
Therefore,a magnetic field cannot change the kinetic energy of a moving charge.
Thus,only an electric field can cause a moving charge to gain energy.

(C) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $\frac{1}{2}mv^2 = eV$.
Here,$e = 1.6 \times 10^{-19} \ C$,$m = 9.1 \times 10^{-31} \ kg$,and $V = 1000 \ V$.
Rearranging for velocity $v$,we get $v = \sqrt{\frac{2eV}{m}}$.
Substituting the values: $v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1000}{9.1 \times 10^{-31}}}$.
$v = \sqrt{\frac{3.2 \times 10^{-16}}{9.1 \times 10^{-31}}} = \sqrt{0.3516 \times 10^{15}} = \sqrt{3.516 \times 10^{14}}$.
$v \approx 1.875 \times 10^7 \ m/s \approx 1.9 \times 10^7 \ m/s$.

(C) For the drop to be stationary,the electric force must balance the gravitational force.
$QE = mg$
Since $E = V/d$,we have $Q(V/d) = mg$,where $Q = ne$.
$n e (V/d) = mg$
$n = \frac{mgd}{Ve}$
Given:
$m = 1.8 \times 10^{-14} \ kg$
$g = 10 \ m/s^2$
$d = 0.90 \ cm = 0.90 \times 10^{-2} \ m$
$V = 2.0 \ kV = 2.0 \times 10^3 \ V$
$e = 1.6 \times 10^{-19} \ C$
Substituting the values:
$n = \frac{1.8 \times 10^{-14} \times 10 \times 0.90 \times 10^{-2}}{2.0 \times 10^3 \times 1.6 \times 10^{-19}}$
$n = \frac{1.62 \times 10^{-14}}{3.2 \times 10^{-16}} = \frac{162}{32} = 5.0625 \approx 5$
Thus,the number of electrons is $5$.

(C) The electron moves with a constant velocity $v_x$ along the $x$-axis.
When a uniform electric field $E$ is applied along the $y$-axis,the electron experiences a constant force $F_y = -eE$ in the negative $y$-direction.
This results in a constant acceleration $a_y = -eE/m$ in the $y$-direction,while the velocity in the $x$-direction remains constant.
Since the motion is uniform in the $x$-direction $(x = v_x t)$ and uniformly accelerated in the $y$-direction $(y = 1/2 a_y t^2)$,the equation of the path is $y = 1/2 (-eE/m) (x/v_x)^2$.
This equation is of the form $y = kx^2$,which represents a parabola.

(A) The kinetic energy $(K)$ gained by an electron accelerated through a potential difference $(V)$ is given by the formula $K = e \cdot V$.
Here,the charge of an electron $e = 1.602 \times 10^{-19} \, C$ and the potential difference $V = 100 \, V$.
Substituting these values into the formula:
$K = (1.602 \times 10^{-19} \, C) \times (100 \, V)$
$K = 1.602 \times 10^{-17} \, J$.
Therefore,the correct option is $A$.

(A) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by the equation: $\frac{1}{2}mv^2 = eV$.
Rearranging for velocity $v$,we get: $v = \sqrt{\frac{2eV}{m}} = \sqrt{2 \left(\frac{e}{m}\right) V}$.
Given values are $V = 200 \ V$ and $\frac{e}{m} = 1.6 \times 10^{11} \ C/kg$.
Substituting these values into the formula:
$v = \sqrt{2 \times (1.6 \times 10^{11}) \times 200}$
$v = \sqrt{6.4 \times 10^{13} \times 10^1} = \sqrt{64 \times 10^{12}}$
$v = 8 \times 10^6 \ m/s$.

(A) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K.E. = eV = \frac{1}{2}mv^2$.
Solving for velocity $v$,we get $v = \sqrt{\frac{2eV}{m}}$.
Given: $e = 1.6 \times 10^{-19} \ C$,$V = 45.5 \ V$,and $m = 9.1 \times 10^{-31} \ kg$.
Substituting the values: $v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 45.5}{9.1 \times 10^{-31}}}$.
$v = \sqrt{\frac{145.6 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{16 \times 10^{12}}$.
$v = 4 \times 10^6 \ m \ s^{-1}$.

(B) The force $F$ acting on a charge $q$ in an electric field $\overrightarrow{E}$ is given by $F = qE$.
According to Newton's second law of motion,$F = ma$,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 $a = \frac{qE}{m}$.
Given the charge $q = 3e$ and mass $m' = 2m$,we substitute these values into the formula:
$a = \frac{(3e)E}{2m} = \frac{3Ee}{2m}$.
Thus,the correct option is $B$.

(B) The kinetic energy gained by the electron is equal to the work done by the electric field: $\frac{1}{2}mv^2 = eV$.
Rearranging for the specific charge $e/m$,we get: $\frac{e}{m} = \frac{v^2}{2V}$.
Substituting the given values $v = 8.4 \times 10^6 \ m/s$ and $V = 200 \ V$:
$\frac{e}{m} = \frac{(8.4 \times 10^6)^2}{2 \times 200} = \frac{70.56 \times 10^{12}}{400} = 0.1764 \times 10^{12} = 1.764 \times 10^{11} \ C/kg$.
Thus,the correct option is $B$.

(A) The force exerted by an electric field on a charged particle is given by $F = qE$. This force can do work on the particle,thereby changing its kinetic energy.
In contrast,the force exerted by a magnetic field on a moving charged particle is given by the Lorentz force $F = q(v \times B)$. Since this force is always perpendicular to the velocity vector $v$ of the particle,the work done by the magnetic field on the particle is always zero $(W = F \cdot d = 0)$.
Therefore,a magnetic field cannot change the kinetic energy of a charged particle. Only the electric field can change the energy of the electron.
Thus,the correct option is $A$.

(B) When an electron is accelerated through a potential difference $V$,the work done by the electric field is equal to the kinetic energy gained by the electron.
The work done is given by $W = eV$.
The kinetic energy gained is $K = \frac{1}{2}mv^2$.
Equating the two,we get $eV = \frac{1}{2}mv^2$.
Rearranging for velocity $v$,we get $v^2 = \frac{2eV}{m}$.
Therefore,the maximum velocity is $v = \sqrt{\frac{2eV}{m}}$.

(C) For the oil drop to be balanced,the electric force must equal the gravitational force acting on it.
$qE = mg$
Given:
Mass $m = 16 \times 10^{-6} \ kg$
Electric field $E = 10^6 \ V/m$
Acceleration due to gravity $g = 10 \ m/s^2$
Rearranging the formula to solve for charge $q$:
$q = \frac{mg}{E}$
Substituting the values:
$q = \frac{16 \times 10^{-6} \times 10}{10^6}$
$q = \frac{16 \times 10^{-5}}{10^6}$
$q = 16 \times 10^{-11} \ C$

(A) The kinetic energy $(K)$ of a charged particle accelerated through a potential difference $(V)$ is given by the formula: $K = qV$.
Here,the charge of an electron is $q = e = 1.602 \times 10^{-19} \ C$ and the potential difference is $V = 100 \ V$.
Therefore,$K = 100 \ eV$.
Converting this to Joules: $K = 100 \times 1.602 \times 10^{-19} \ J = 1.602 \times 10^{-17} \ J$.

(D) Initially,the particle $q$ is at an infinite distance from $Q$,so the initial potential energy $P.E._i = 0$.
Case $1$: Initial velocity is $v$ and the closest distance is $r$.
Applying the law of conservation of energy: $P.E._i + K.E._i = P.E._f + K.E._f$
$0 + \frac{1}{2}mv^2 = \frac{kQq}{r} + 0$
$\frac{1}{2}mv^2 = \frac{kQq}{r} \quad \dots (1)$
Case $2$: Initial velocity is $2v$ and let the new closest distance be $r'$.
Applying the law of conservation of energy: $P.E._i + K.E._i = P.E._f + K.E._f$
$0 + \frac{1}{2}m(2v)^2 = \frac{kQq}{r'} + 0$
$\frac{1}{2}m(4v^2) = \frac{kQq}{r'} \quad \dots (2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{\frac{1}{2}mv^2}{\frac{1}{2}m(4v^2)} = \frac{\frac{kQq}{r}}{\frac{kQq}{r'}}$
$\frac{1}{4} = \frac{r'}{r}$
$r' = \frac{r}{4}$

(C) An electron carries a negative charge $(q = -e)$.
When it enters an electric field $\vec{E}$ with a velocity $\vec{v}$ parallel to the field lines,the electric force acting on it is given by $\vec{F} = q\vec{E} = -e\vec{E}$.
Since the force is in the opposite direction to the electric field lines,the force acts in the direction opposite to the electron's velocity.
This force causes a retardation (deceleration) in the electron's motion.
Therefore,the velocity of the electron will decrease as it moves through the field.

(B) The force $F$ acting on an electron in an electric field $E$ is given by $F = eE$.
According to Newton's second law,$F = ma$,where $m$ is the mass of the electron.
Equating the two,we get $ma = eE$,which implies $a = \frac{eE}{m}$.
Given: $E = 9.1 \times 10^6 \ N/C$,$e = 1.6 \times 10^{-19} \ C$,and $m = 9.1 \times 10^{-31} \ kg$.
Substituting the values: $a = \frac{(1.6 \times 10^{-19}) \times (9.1 \times 10^6)}{9.1 \times 10^{-31}}$.
$a = 1.6 \times 10^{-19+6+31} = 1.6 \times 10^{18} \ m/s^2$.

(B) For the water drop to remain suspended in air,the electric force must balance the gravitational force.
$F_e = mg$
$eE = mg$
$E = \frac{mg}{e}$
Here,$m = \text{volume} \times \text{density} = \frac{4}{3} \pi r^3 \rho$.
Given: $r = 10^{-5} \, cm = 10^{-7} \, m$,$\rho = 1000 \, kg/m^3$,$e = 1.6 \times 10^{-19} \, C$,$g = 10 \, m/s^2$.
$m = \frac{4}{3} \times 3.14 \times (10^{-7})^3 \times 1000 = \frac{4}{3} \times 3.14 \times 10^{-18} \, kg$.
$E = \frac{\frac{4}{3} \times 3.14 \times 10^{-18} \times 10}{1.6 \times 10^{-19}} \approx 261.6 \, N/C$.
Rounding to the nearest provided option,the correct value is $260 \, N/C$.

(C) The force on an electron in an electric field $E$ is given by $F_e = eE$. According to Newton's second law,$m_e a_e = eE$,so $a_e = \frac{eE}{m_e}$.
The force on a proton in the same electric field $E$ is $F_p = eE$. According to Newton's second law,$m_p a_p = eE$,so $a_p = \frac{eE}{m_p}$.
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) When a charged particle enters a uniform electric field $E$ perpendicular to its direction of motion,it experiences a force $F = qE$,resulting in an acceleration $a = \frac{qE}{m}$.
Assuming the particle enters with velocity $v$ and travels a horizontal distance $x$,the time taken is $t = \frac{x}{v}$.
The vertical deflection $y$ is given by $y = \frac{1}{2}at^2 = \frac{1}{2} \left( \frac{qE}{m} \right) \left( \frac{x}{v} \right)^2$.
Since $y \propto \frac{q}{m}$,the particle with the greatest vertical deflection has the highest charge-to-mass ratio.
In the figure,particle $(3)$ shows the maximum deflection,therefore,particle $(3)$ has the highest charge-to-mass ratio.

(A) The electron enters the electric field perpendicular to the field lines. The horizontal velocity $u$ remains constant as there is no force in the horizontal direction.
The time taken to travel a horizontal distance $x$ is $t = \frac{x}{u}$.
The vertical force on the electron is $F = qE$,so the vertical acceleration is $a = \frac{qE}{m}$.
The vertical deflection $y$ is given by $y = \frac{1}{2}at^2 = \frac{1}{2} \left( \frac{qE}{m} \right) \left( \frac{x}{u} \right)^2$.
Given: $E = 3200 \ V/m$,$x = 0.10 \ m$,$u = 4 \times 10^7 \ m/s$,$q = 1.6 \times 10^{-19} \ C$,$m = 9.1 \times 10^{-31} \ kg$.
Substituting the values:
$y = \frac{1}{2} \times \frac{1.6 \times 10^{-19} \times 3200}{9.1 \times 10^{-31}} \times \frac{(0.1)^2}{(4 \times 10^7)^2}$
$y = \frac{1}{2} \times \frac{5.12 \times 10^{-16}}{9.1 \times 10^{-31}} \times \frac{0.01}{16 \times 10^{14}}$
$y = \frac{5.12 \times 10^{-16} \times 10^{-2}}{2 \times 9.1 \times 10^{-31} \times 16 \times 10^{14}}$
$y \approx 0.001758 \ m \approx 1.76 \ mm$.