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: ____________________

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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 $q$ in an electric field $E$.

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Section B – Numerical Problems

(Use: $e = 1.6 \times 10^{-19} \, C$,
Mass of electron $=9.1\times {10}^{-31}kg$ 

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

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11. An electron enters a uniform electric field of $2 \times 10^4 \, N/C$. Calculate its acceleration.

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

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

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

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

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

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

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

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

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

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✅ 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 $q$, electric field $E$, and mass $m$.

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

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

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

9. Direction of deflection reverses.

10. $F = qE$
    

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Section B Answers

$$a = \frac{qE}{m} = \frac{(1.6 \times 10^{-19})(2 \times 10^4)}{9.1 \times 10^{-31}} \approx 3.52 \times 10^{15} \, m/s^2$$

$$F = qE = (1.6 \times 10^{-19})(5 \times 10^3) = 8 \times 10^{-16} \, N$$

$$a = \frac{qE}{m} = \frac{(2 \times 10^{-6})(500)}{1 \times 10^{-3}} = 1000 \, m/s^2$$

$$t = \frac{\text{length}}{\text{velocity}} = \frac{0.05}{4 \times 10^6} = 1.25 \times 10^{-8} \, s$$

$$a = 3.52 \times 10^{15} \, m/s^2$$
$$y = \frac{1}{2} a t^2 \approx 0.275 \, mm$$

$$a = \frac{qE}{m} = \frac{(1.6 \times 10^{-19})(10^4)}{1.67 \times 10^{-27}} \approx 9.58 \times 10^{11} \, m/s^2$$
$$\Delta v = at = (9.58 \times 10^{11})(2 \times 10^{-6}) \approx 1.92 \times 10^6 \, m/s$$

17. Parabolic path.

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

19. Acceleration doubles (since $a = qE/m$).

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