Download App

Electric Charges and Fields — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields4 Marks
Question
When a charged particle is placed in an electric field, it experiences an electrical force. If this is the only force on the particle, it must be the net force. The net force will cause the particle to accelerate according to Newton's second law. So $\vec{\text{F}}_\text{e}=\text{q}\vec{\text{E}}=\text{m}\vec{\text{a}}$

If $\vec{\text{E}}$ is uniform, then $\vec{\text{a}}$ is constant and $\vec{\text{a}}=\text{q}\vec{\text{E}}\text{/ m.}$ If the particle has a positive charge, its acceleration is in the direction of the field. If the particle has a negative charge, its acceleration is in the direction opposite to the electric field. Since the acceleration is constant, the kinematic equations can be used.

  1. An electron of mass m, charge e falls through a distance h metre in a uniform electric field E. Then time of fall,
  1. $\text{t}=\sqrt{\frac{\text{2hm}}{\text{eE}}}$

  2. $\text{t}=\frac{\text{2hm}}{\text{eE}}$

  3. $\text{t}=\sqrt{\frac{\text{2eE}}{\text{hm}}}$

  4. $\text{t}=\frac{\text{2eE}}{\text{hm}}$

  1. An electron moving with a constant velocity v along X-axis enters a uniform electric field applied along Y-axis. Then the electron moves:
  1. With uniform acceleration along Y-axis
  2. Without any acceleration along Y-axis
  3. In a trajectory represented as y = ax2
  4. In a trajectory represented as y = ax
  1. Two equal and opposite charges of masses m1 and m2 are accelerated in an uniform electric field through the same distance. What is the ratio of their accelerations if their ratio of masses is $\frac{\text{m}_1}{\text{m}_2}=0.5?$
  1. $\frac{\text{a}_1}{\text{a}_2}=2$

  2. $\frac{\text{a}_1}{\text{a}_2}=0.5$

  3. $\frac{\text{a}_1}{\text{a}_2}=3$

  4. $\frac{\text{a}_1}{\text{a}_2}=1$

  1. A particle of mass m carrying charge q is kept at rest in a uniform electric field E and then released. The kinetic energy gained by the particle, when it moves through a distance y is:
  1. $\frac{1}{2}\text{qEy}^2$
  2. qEy
  3. qEy2
  4. qE2y
  1. A charged particle is free to move in an electric field. It will travel:
  1. Always along a line of force.
  2. Along a line of force, if its initial velocity is zero.
  3. Along a line of force, if it has some initial velocity in the direction of an acute angle with the line of force.
  4. None of these.
✓

Answer

  1. (a) $\text{t}=\sqrt{\frac{\text{2hm}}{\text{eE}}}$

Explanation:

From Newton's law,

$\text{F}=\text{m}\vec{\text{a}}$

Or $\text{qE}=\text{m}\vec{\text{a}}$

$\Rightarrow\text{a}=\frac{\text{qE}}{\text{m}}=\frac{\text{eE}}{\text{m}}$

Using, $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

$\therefore\text{ h}=0+\frac{1}{2}\times\frac{\text{eE}}{\text{m}}\text{t}^2$

$\Rightarrow\text{t}=\sqrt{\frac{\text{2hm}}{\text{eE}}}$

  1. (c) In a trajectory represented as y = ax2
  2. (b) $\frac{\text{a}_1}{\text{a}_2}=0.5$

Explanation:

Force is same in magnitude for both.

$\therefore$ m1a1 = m2a2;

$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{m}_2}{\text{m}_1}=\frac{1}{0.5}=2$

  1. (b) qEy

Explanation:

Here, u = 0;

$\text{a}=\frac{\text{qE}}{\text{m}};\text{ s}=\text{y}$

Using, v2 - u2 = 2as

$\Rightarrow\text{v}^2=2\frac{\text{qE}}{\text{m}}\text{y}$

$\therefore\text{ K.E.}=\frac{\text{1}}{\text{2}}\text{my}^2=\text{qEy}$

  1. (b) Along a line of force, if its initial velocity is zero.

Explanation:

If charge particle is put at rest in electric field, then it will move along line of force.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks Questions" from Electric Charges and Fields→Electric Charges and Fields — all question types→Physics STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

When a fat person tries to touch his toes, keeping the legs straight, he generally falls. Explain with reference to figure.

→
When a tyre bursts, the air coming out is cooler than the surrounding air. Explain.
→
Two point charges $+3.2 \times 10^{-19}$ Coulomb and $-3.2 \times 10^{-19}$ Coulomb are kept at a distance of 2.4 $\times 10^{-10}$ meters from each other. This dipole is placed at an angle of $30^{\circ}$ to the field direction in a uniform field of intensity of $3 \times 10^4$ Newton/Coulomb. Find the value of the moment of the couple acting on it. What will be the maximum value of this torque?
→
The charge density of two uniformly charged parallel infinite plane sheets ' 1 ' and ' 2 ' are $+\sigma$ and $-2 \sigma$ respectively. Find the magnitude and direction of the net electric field (i) at any point between these two sheets and (ii) at any point outisde these two sheets but near sheet ' 1 '.
→

In 1911, Rutherford, along with his assistants, H. Geiger and E. Marsden, performed the alpha particle scattering experiment. H. Geiger and E. Marsden took radioactive source $(^{214}_{83}\text{Bi})$ for $\alpha$- particles. A collimated beam of $\alpha$-particles of energy 5.5 MeV was allowed to fall on 2.1 × 10-7 m thick gold foil. The $\alpha$-particles were observed through a rotatable detector consisting of a Zinc sulphide screen and microscope. It was found that CL-particles got scattered. These scattered $\alpha$-particles produced scintillations on the zinc sulphide screen. Observations of this experiment are as follows?

Most of the $\alpha$-particles passed through the foil without deflection.

Only about 0.14% of the incident $\alpha$-particles scattered by more than 1º

Only about one $\alpha$-particle in every 8000 $\alpha$-particles deflected by more than 90º

These observations led to many arguments and conclusions which laid down the structure of the nuclear model of an atom.

  1. Rutherford's atomic model can be visualised as.

  1. Gold foil used in Geiger-Marsden experiment is about 10-8 m thick. This ensures.
  1. Gold foil's gravitational pull is small or possible.
  2. Gold foil is deflected when $\alpha$-particle stream is not incident centrally over it.
  3. Gold foil provides no resistance to passage of $\alpha$-particles.
  4. Most $\alpha$-particle will not suffer more than 1º scattering during passage through gold foil.
  1. In Geiger-Marsden scattering experiment, the trajectory traced by an $\alpha$-particle depends on.
  1. Number of collision.
  2. Number of scattered $\alpha$- particles.
  3. Impact parameter.
  4. None of these.
  1. In the Geiger-Marsden scattering experiment, in case of head-on collision, the impact parameter should be.
  1. Maximum
  2. Minimum
  3. Infinite
  4. zero
  1. The fact only a small fraction of the number of incident particles rebound back in Rutherford scattering indicates that.
  1. Number of $\alpha$-particles undergoing head-on-collision is small.
  2. Mass of the atom is concentrated in a small volume.
  3. Mass of the atom is concentrated in a large volume.
  4. Both (a) and (b).
→
The moon rotates about the earth in such a way that only one hemisphere of the moon faces the earth. Can we ever see the ''other face'' of the moon from the earth? Can a person on the moon ever see all the faces of the earth?

→
Two fat astronauts each of mass 120kg are travelling in a closed spaceship moving at a speed of 15km/s in the outer space far removed from all other material objects. The total mass of the spaceship and its contents including the astronauts is 660kg. If the astronauts do slimming exercise and thereby reduce their masses to 90kg each, with what velocity will the spaceship move?
→
When a glass capillary tube is dipped at one end in water, water rises in the tube. The gravitational potential energy is thus increased. Is it a violation of conservation of energy?
→
On a winter day, the outside temperature is 0°C and relative humidity 40%. The air from outside comes into a room and is heated to 20°C. What is the relative humidity in the room? The saturation vapour pressure at 0°C is 4.6mm of mercury and at 20°C it is 18mm of mercury.
→
Show that the magnetic field at a point due to a magnetic dipole is perpendicular to the magnetic axis if the line joining the point with the centre of the dipole makes an angle of $\tan^{-1}(\sqrt{2})$ with the magnetic axis.
→