Showing posts with label 12P01 Electric Charges and Fields. Show all posts
Showing posts with label 12P01 Electric Charges and Fields. Show all posts

WS Level 1 Charged Particle in a Uniform Electric Field

 

From the handbook


📘 WORKSHEET  Level 1

Charged Particle in a Uniform Electric Field

Name: ____________________

Class: ____________________

Date: ____________________


Section A – Conceptual Questions

  1. What type of motion does a charged particle undergo when it enters a uniform electric field perpendicular to its initial velocity?

  2. Why is the motion of a charged particle in a uniform electric field comparable to projectile motion?

  3. A positively charged particle enters a uniform electric field directed to the right. In which direction does it accelerate?

  4. A negatively charged particle enters the same field as in Q3. In which direction does it accelerate?

  5. On what factors does the acceleration of a charged particle in a uniform electric field depend?

  6. If the electric field strength is doubled, what happens to the force acting on the particle?

  7. Can a charged particle move undeflected inside a uniform electric field? If yes, under what condition?

  8. How does increasing the mass of the particle affect its deflection in the field?

  9. What happens to the trajectory if the sign of charge is reversed?

  10. Write the expression for the force experienced by a charge qq in an electric field EE.


Section B – Numerical Problems

(Use: e=1.6×1019e = 1.6 \times 10^{-19} \, C,
Mass of electron =9.1×1031kg 

Mass of proton =1.67×1027kg= 1.67 \times 10^{-27} \, kg)


  1. An electron enters a uniform electric field of 2×104N/C2 \times 10^4 \, N/C. Calculate its acceleration.

  2. A proton enters a uniform electric field of 5×103N/C5 \times 10^3 \, N/C. Find the force acting on it.

  3. A particle of charge 2×106C2 \times 10^{-6} \, C and mass 1×103kg1 \times 10^{-3} \, kg enters a uniform electric field of 500 N/C.
    Find its acceleration.

  4. An electron enters horizontally between two parallel plates separated by 2 cm. Electric field between plates is 1×104N/C1 \times 10^4 \, N/C. If its horizontal velocity is 4×106m/s4 \times 10^6 \, m/s, find the time taken to cross a plate of length 5 cm.

  5. For Q14, calculate the vertical deflection of the electron.

  6. A proton enters a uniform electric field parallel to its velocity. If E=104N/CE = 10^4 \, N/C, find its change in velocity after 2μs2 \, \mu s.

  7. An electron enters a uniform electric field perpendicular to its velocity.
    If the field region length is 5 cm and E=3×104N/CE = 3 \times 10^4 \, N/C, determine the nature of the trajectory.

  8. A charged particle enters a uniform electric field and emerges without deflection.
    What does this indicate about the direction of motion?

  9. If the charge on a particle is doubled while keeping mass constant, how does acceleration change?

  10. An electron enters an upward electric field with horizontal velocity.
    In which direction will it deflect? Why?


✅ ANSWERS

Section A Answers

  1. Parabolic motion.

  2. Because acceleration is constant (like gravity in projectile motion).

  3. To the right (along the field).

  4. To the left (opposite to the field).

  5. Depends on charge qq, electric field EE, and mass mm.

  6. Force doubles (since F=qEF = qE).

  7. Yes, if velocity is parallel or antiparallel to the field.

  8. Deflection decreases (since a=qE/ma = qE/m).

  9. Direction of deflection reverses.

  10. F=qEF = qE


Section B Answers

a=qEm=(1.6×1019)(2×104)9.1×10313.52×1015m/s2a = \frac{qE}{m} = \frac{(1.6 \times 10^{-19})(2 \times 10^4)}{9.1 \times 10^{-31}} \approx 3.52 \times 10^{15} \, m/s^2
F=qE=(1.6×1019)(5×103)=8×1016NF = qE = (1.6 \times 10^{-19})(5 \times 10^3) = 8 \times 10^{-16} \, N
a=qEm=(2×106)(500)1×103=1000m/s2a = \frac{qE}{m} = \frac{(2 \times 10^{-6})(500)}{1 \times 10^{-3}} = 1000 \, m/s^2
t=lengthvelocity=0.054×106=1.25×108st = \frac{\text{length}}{\text{velocity}} = \frac{0.05}{4 \times 10^6} = 1.25 \times 10^{-8} \, s
a=3.52×1015m/s2a = 3.52 \times 10^{15} \, m/s^2
y=12at20.275mmy = \frac{1}{2} a t^2 \approx 0.275 \, mm
a=qEm=(1.6×1019)(104)1.67×10279.58×1011m/s2a = \frac{qE}{m} = \frac{(1.6 \times 10^{-19})(10^4)}{1.67 \times 10^{-27}} \approx 9.58 \times 10^{11} \, m/s^2
Δv=at=(9.58×1011)(2×106)1.92×106m/s\Delta v = at = (9.58 \times 10^{11})(2 \times 10^{-6}) \approx 1.92 \times 10^6 \, m/s
  1. Parabolic path.

  2. Velocity is parallel (or antiparallel) to the electric field.

  3. Acceleration doubles (since a=qE/ma = qE/m).

  4. It deflects downward (opposite to field direction) because electron is negatively charged.

A sphere encloses an electric dipole with charges ± 3 × 10^−6 C. What is the total electric flux across the sphere ?

The electric field in a certain region is acting radially outward and is given by E = Aa. A charge contained in a sphere of radius ’ a ’ centered at the origin of the field, will be given by ?

What is the flux through a cube of side ’ a ’ if a point charge of q is at one of its corner ?

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will ?

A square surface of side L meter in the plane of the paper is placed in a uniform electric field E (volt/m) acting along the same plane at an angle θ with the horizontal side of the square as shown in Figure. The electric flux linked to the surface, in units of volt. m, is ?

A hollow cylinder has a charge q coulomb within it. If φ is the electric flux in units of voltmeter associated with the curved surface B, the flux linked with the plane surface A in units of voltmeter will be ?

A square surface of side L metres is in the plane of the paper. A uniform electric field E (volt/m ), also in the plane of the paper, is limited only to the lower half of the square surface (see figure). The electric flux in SI units associated with the surface is ?

A charge QμC is placed at the centre of a cube, the flux coming out from any surface will be ?

A charge Q is placed at the comer of a cube. The electric flux through all the six faces of the cube is ?

A point charge +q is placed at mid point of a cube of side ’L’. The electric flux emerging from the cube is ?

A spherical conductor of radius 10 cm has a charge of 3.2 × 10−7C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere ?

Two parallel infinite line charges with linear charge densities +λC/m and −λC/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges ?

A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at ?

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is ?

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE , its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively ?