WS Level 2 Charged Particle in a Uniform Electric Field

 Here is a Higher-Difficulty Worksheet (JEE/NEET Level) on:

📘 Charged Particle in a Uniform Electric Field

(Advanced Problems – JEE / NEET Level)

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🔷 Section A – Conceptual & Analytical

1. A particle of charge $q$ and mass $m$ enters a uniform electric field $\vec{E}$ with velocity $\vec{u}$ making an angle $\alpha$ with the field. Derive the equation of its trajectory.

2. Show that the path of a charged particle in a uniform electric field is a parabola. Obtain the expression for its maximum displacement from the initial line of motion.

3. A charged particle enters a uniform electric field perpendicular to its velocity.
   Prove that the angle of deflection $\theta$ at the exit is:

   $$\tan\theta = \frac{qEL}{m u^2}$$

   where $L$ is the length of the field region.

4. Under what condition will a charged particle move in a straight line inside a uniform electric field? Discuss all possible cases.

5. A particle enters a region of uniform electric field and emerges with the same speed but different direction. What does this indicate about the direction of the electric field relative to velocity?

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🔷 Section B – Numerical / Multi-Concept 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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6. Deflection Between Parallel Plates

An electron enters horizontally between two parallel plates of length 4 cm.
Electric field between plates is $2 \times 10^4 \, N/C$. Initial velocity $= 5 \times 10^6 \, m/s$.

Find:

a) Time of flight inside the field
b) Vertical deflection
c) Angle of emergence

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7. Required Electric Field for Zero Deflection

An electron enters a region with velocity $3 \times 10^6 \, m/s$. A uniform electric field is applied such that the electron emerges undeflected.

Find the direction of electric field required and explain physically.

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8. Minimum Speed Condition

A proton enters a uniform electric field of magnitude $5 \times 10^3 \, N/C$ perpendicular to the field.
If the field region length is 10 cm, what minimum initial speed must it have so that its deflection does not exceed 1 mm?

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9. Energy Approach Problem

An electron enters a uniform electric field parallel to its velocity.
If it travels 2 cm inside a field of $10^4 \, N/C$, find:

a) Work done by the field
b) Increase in kinetic energy
c) Final velocity

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10. Time of Flight with Inclined Entry

A particle of charge $2\mu C$ and mass $2 \times 10^{-3} \, kg$ enters a uniform electric field at 30° to the field direction.
Field strength = 500 N/C.
Initial speed = 20 m/s.

Find:

a) Time after which velocity becomes perpendicular to the electric field
b) Its speed at that instant

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11. Ratio Problem (JEE Type)

An electron and proton enter the same uniform electric field with equal initial velocities perpendicular to the field.

Find the ratio of:

a) Their accelerations
b) Their time of flight (if field region length is same)
c) Their vertical deflections

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12. Angle of Emergence Derivation

A charged particle enters a uniform electric field perpendicular to its velocity.
Derive an expression for angle of emergence in terms of:

$$\theta = \tan^{-1}\left(\frac{qEL}{mu^2}\right)$$

and discuss limiting cases:

• $u\to \mathrm{\infty }$

• $m \to \infty$
  

• $E \to 0$
  

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🔷 Section C – Assertion & Reason (NEET Level)

Assertion: A charged particle moving parallel to a uniform electric field always moves with constant velocity.
Reason: Electric force acts perpendicular to velocity.

Assertion: The trajectory of a charged particle in uniform electric field is always parabolic.
Reason: Acceleration due to electric field is constant.

Assertion: Deflection is inversely proportional to mass of particle.
Reason: Acceleration is inversely proportional to mass.

🔷 SECTION A – Analytical Solutions

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1. Derive Equation of Trajectory (General Case)

Let electric field $\vec{E}$ be along +y direction.
Particle enters with velocity $\vec{u}$ making angle $\alpha$ with x-axis.

Step 1: Resolve initial velocity

$$u_x = u\cos\alpha$$
$$u_y = u\sin\alpha$$

Step 2: Acceleration

Force:

$$F = qE$$
$$a = \frac{qE}{m}$$

Acceleration only in y-direction.

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Step 3: Equations of motion

Horizontal motion:

$$x = u\cos\alpha \; t$$

Vertical motion:

$$y = u\sin\alpha \; t + \frac{1}{2}\frac{qE}{m}t^2$$

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Step 4: Eliminate $t$

$$t = \frac{x}{u\cos\alpha}$$

Substitute:

$$y = x\tan\alpha + \frac{qE}{2m u^2 \cos^2\alpha} x^2$$

✅ This is equation of a parabola.

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2. Maximum Displacement

If particle enters perpendicular to field ($\alpha=0$):

$$y = \frac{qE}{2m u^2} x^2$$

At exit $x=L$:

$$y_{max} = \frac{qE L^2}{2m u^2}$$

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3. Angle of Deflection

Time inside field:

$$t = \frac{L}{u}$$Vertical velocity gained:

$$v_y = at = \frac{qE}{m}\frac{L}{u}$$Horizontal velocity remains u.

$$\tan\theta = \frac{v_y}{u}$$ $$\tan\theta = \frac{qEL}{m u^2}$$

✔ Proven.

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4. Straight Line Motion Condition

Straight line motion if:

• $q=0$
  

• $E=0$
  

• Velocity parallel or antiparallel to $E$
  

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5. Same Speed but Different Direction

If speed unchanged:

$$W = qEd = 0$$

⇒ Electric field perpendicular to velocity.

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🔷 SECTION B – Numerical Solutions

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6. Electron Between Plates

Given:

$$E = 2\times10^4$$
$$L = 0.04 m$$
$$u = 5\times10^6$$

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(a) Time

$$t = \frac{L}{u} = \frac{0.04}{5\times10^6} = 8\times10^{-9} s$$

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(b) Acceleration

$$a = \frac{qE}{m}$$ $$= \frac{1.6\times10^{-19}\times2\times10^4}{9.1\times10^{-31}}$$ $$a = 3.52\times10^{15} m/s^2$$

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(c) Deflection

$$y = \frac{1}{2}at^2$$
$$= 0.5 \times 3.52\times10^{15} \times (8\times10^{-9})^2$$
$$y = 0.113 mm$$

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(d) Angle

$$\tan\theta = \frac{qEL}{mu^2}$$ $$\theta \approx 0.056 rad$$

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8. Minimum Speed Condition

Allowed deflection:

$$y = 1 mm = 10^{-3} m$$

Use:

$$y = \frac{qE L^2}{2m u^2}$$

Solve for $u$:

$$u = \sqrt{\frac{qEL^2}{2my}}$$

Substitute values:

$$u \approx 3.9\times10^4 m/s$$

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9. Energy Method

Distance = 0.02 m

Work done:

$$W = qEd$$
$$= 1.6\times10^{-19}\times10^4\times0.02$$
$$= 3.2\times10^{-17} J$$

Increase in KE = Work done

$$\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = 3.2\times10^{-17}$$

Solve for $v$.

$$v \approx 2.65\times10^6 m/s$$

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10. Velocity Perpendicular Condition

Velocity becomes perpendicular when:

$$v_y = 0$$

Use:

$$v_y = u\sin30° - at$$

Solve for $t$:

$$t = \frac{u\sin30°}{a}$$

Substitute:

$$a = \frac{qE}{m} = \frac{2\times10^{-6}\times500}{2\times10^{-3}} = 0.5$$
$$t = \frac{20\times0.5}{0.5} = 20 s$$

Speed at that instant:

$$v = u\cos30° = 17.32 m/s$$

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11. Electron vs Proton Ratio

(a) Acceleration

$$a = \frac{qE}{m}$$ $$\frac{a_e}{a_p} = \frac{m_p}{m_e}$$ $$= 1836$$

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(b) Time of flight

Same (depends only on horizontal velocity).

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(c) Deflection

$$\frac{y_e}{y_p} = \frac{m_p}{m_e}$$ $$= 1836$$

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🔷 SECTION C – Assertion & Reason

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13.

Assertion ❌ False
Reason ❌ False

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14.

Assertion ❌ False (only if perpendicular entry)
Reason ✔ True

Correct option: Reason correct but assertion not always true.

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15.

Assertion ✔ True
Reason ✔ True

Correct option: Both true and reason correct explanation.