WS Level 3 Charged Particle in a Uniform Electric Field

 

Excellent 😎🔥
Now we move to JEE Advanced – Conceptual + Multi-Concept + Tricky Problems
Topic: Charged Particle in Uniform Electric Field

These are high-difficulty, multi-step, conceptual trap style problems.

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🚀 EXTREMELY TOUGH – JEE ADVANCED LEVEL

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🔴 Problem 1: Variable Entry + Constrained Exit Condition

A particle of charge $q$ and mass $m$ enters a uniform electric field $\stackrel{}{E}$ (directed vertically upward) with velocity $u$ at angle $\theta$ above horizontal.

Length of field region = $L$.

The particle just grazes the upper plate at the exit.

Find:

1. The required value of $\theta$

2. The maximum height reached inside the field

3. Condition for which particle just exits horizontally

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🔴 Problem 2: Electric Field Switching Midway

A charged particle enters a uniform electric field $E$ perpendicular to its velocity.

After traveling half the length of the plates, the electric field is suddenly reversed.

Find:

1. Net deflection at exit

2. Angle of emergence

3. Condition for zero net deflection

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🔴 Problem 3: Drift Cancellation Problem (Trap Question)

A particle enters a region where uniform electric field $E$ exists vertically upward.

It enters horizontally.

Inside the field, gravity also acts downward.

Find:

1. Condition for straight line motion

2. Nature of trajectory if $q\mathrm{E }> m\mathrm{g }$

3. Nature if $q\mathrm{E }< mg$

🔴 Problem 4: Time Minimization (Calculus Based)

A charged particle must cross a field region of length $L$.

It enters at adjustable angle $\theta$.

Find the angle $\theta$ that minimizes:

1. Time spent inside field

2. Vertical deflection

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🔴 Problem 5: Energy + Kinematics Combined

A particle enters parallel to electric field with speed $u$.

After traveling distance $d$, its velocity doubles.

Find:

1. Electric field strength

2. Time taken

3. Work done by field

4. Potential difference crossed

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🔴 Problem 6: Re-entry Condition

A particle enters perpendicular to field.

After exiting, it re-enters the same field region due to deflection.

Find condition for re-entry in terms of:

$$\frac{qEL}{mu^2}$$

🔴 Problem 7: Critical Speed for No Plate Collision

Plates separated by distance $d$.

Field = $E$

Length = $L$

Find minimum speed required so particle does not strike either plate.

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🔴 Problem 8: Frame of Reference Trap

A particle enters a uniform electric field.

Observer moves horizontally with speed $v$.

Find trajectory equation in moving frame.

Is it still parabolic?

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🔴 Problem 9: Multi-Particle Comparison

Electron, proton, and alpha particle enter same electric field with same kinetic energy.

Find ratio of:

1. Maximum deflections

2. Angles of emergence

3. Time of flight

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🔴 Problem 10: Combined Projectile + Electric Field

A charged particle is projected upward at angle $\theta$ in region where:

• Electric field acts upward

• Gravity acts downward

Find condition under which motion becomes purely horizontal at some point.

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⚠️ These Problems Test:

• Deep kinematics

• Energy methods

• Frame transformation

• Limiting cases

• Ratio reasoning

• Hidden assumptions

• Sign conventions

• Calculus optimization

Perfect 👍
Here are JEE Advanced–style hints only (no direct solving).
Use them like exam guidance — minimal but directional.

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🔴 Problem 1: Variable Entry + Grazing Condition

Hints:

• Resolve velocity into components along and perpendicular to field.

• Write trajectory equation:

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

• “Just grazes” ⇒ displacement at $x=L$ equals plate separation.

• For horizontal exit: set $v_y = 0$ at $x=L$.

• Use:

  $$v_y = u\sin\theta + \frac{qE}{m}t$$

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🔴 Problem 2: Field Reversal Midway

Hints:

• Divide motion into two equal time intervals.

• First half: acceleration = $+a$

• Second half: acceleration = $-a$

• Horizontal velocity constant throughout.

• Compare vertical velocity gained in first half vs lost in second.

• Symmetry check: what happens if time intervals equal?

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🔴 Problem 3: Gravity + Electric Field

Hints:

• Net vertical acceleration:

  $$a_{net} = \frac{qE}{m} - g$$
• Straight line motion ⇒ $a_{net} = 0$
• Compare magnitudes of $qE$ and $mg$.
• If unequal, motion still parabolic — but curvature direction changes.

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🔴 Problem 4: Time Minimization (Calculus)

Hints:

• Time inside field depends only on horizontal component: t=L/(ucosθ)

  $$\[t = \frac{L}{u\cos\theta}\]$$

• Minimize $t$ ⇒ maximize $\cos\theta$

• For deflection minimization:

  $$y \propto \frac{1}{u^2\cos^2\theta}$$

• Take derivative wrt $\theta$ and set = 0.

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🔴 Problem 5: Velocity Doubles

Hints:

• Use energy conservation:

  $$qEd = \frac{1}{2}m(4u^2 - u^2)$$

• Electric field constant ⇒ uniform acceleration.

• Use:

  $$v^2 = u^2 + 2ad$$

• Time from:

  $$v = u + at$$

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🔴 Problem 6: Re-entry Condition

Hints:

• Find angle of emergence:

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

• After exit, motion is straight line.

• Check whether projected trajectory intersects entry boundary again.

• Re-entry requires sufficient vertical velocity.

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🔴 Problem 7: No Plate Collision

Hints:

• Maximum vertical deflection:

  $$y = \frac{qEL^2}{2mu^2}$$

• Set $y \leq d/2$

• Solve inequality for minimum $u$.

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🔴 Problem 8: Moving Frame

Hints:

• In moving frame:

  $$u'_x = u_x - v$$

• Vertical acceleration unchanged.

• Trajectory equation form unchanged.

• Parabola remains parabola (Galilean transformation).

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🔴 Problem 9: Same KE Comparison

Hints:

• Same KE ⇒

  $$\frac{1}{2}mv^2 = constant$$

• So $v \propto \frac{1}{\sqrt{m}}$
  

• Deflection:

• $$y \propto \frac{q}{m}\frac{1}{v^2}$$

• Carefully substitute $v^2$.

Trap: mass cancels partially.

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🔴 Problem 10: Electric Field + Gravity

Hints:

• Net acceleration:

  $$a = \frac{qE}{m} - g$$

• For horizontal velocity condition:

  $$vy=0v_y = 0$$

• Use:

  $$v_y = u\sin\theta - at$$

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🧠 JEE Advanced Strategy Reminder

• Always separate horizontal and vertical motion.

• Use energy method when motion is along field.

• Use kinematics when motion is perpendicular.

• Watch signs carefully.

• Check limiting cases (E→0, m→∞, u→∞).

🔴 PROBLEM 1

Variable Entry + Grazing + Horizontal Exit

A particle enters a uniform electric field $\vec{E}$ (upward).
Initial speed = $u$, angle = $\theta$.
Length of plates = $L$.

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Step 1: Resolve motion

Horizontal:

$$u_x = u\cos\theta$$

Vertical:

$$u_y = u\sin\theta$$

Acceleration:

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

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Step 2: Time inside field

$$t = \frac{L}{u\cos\theta}$$

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Step 3: Vertical displacement at exit

$$y = u\sin\theta \cdot t + \frac{1}{2}at^2$$

Substitute $t$:

$$y = u\sin\theta \frac{L}{u\cos\theta} + \frac{1}{2}\frac{qE}{m}\left(\frac{L}{u\cos\theta}\right)^2$$ $$y = L\tan\theta + \frac{qEL^2}{2mu^2\cos^2\theta}$$

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(1) Condition for grazing upper plate

If plate separation = $d$

$$d = L\tan\theta + \frac{qEL^2}{2mu^2\cos^2\theta}$$

This is required equation for $\theta$.

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(2) Maximum height inside field

Maximum height occurs when:

$$v_y = 0$$ $$v_y = u\sin\theta + at$$

Set = 0:

$$t_{max} = -\frac{u\sin\theta}{a}$$

Substitute into vertical displacement formula.

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(3) Condition for horizontal exit

Horizontal exit ⇒ $v_y = 0$ at $t = L/(u\cos\theta)$

$$u\sin\theta + \frac{qE}{m}\frac{L}{u\cos\theta} = 0$$ $$\tan\theta = -\frac{qEL}{mu^2}$$

This is exact condition.

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🔴 PROBLEM 2

Field Reversal at Midpoint

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Step 1: Total time

$$t = \frac{L}{u}$$

Half time:

$$t_1 = \frac{L}{2u}$$

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Step 2: First half motion

Acceleration = $+a$

Velocity gained:

$$v_{y1} = a t_1$$

Displacement:

$$y_1 = \frac{1}{2}a t_1^2$$

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Step 3: Second half

Acceleration = $-a$

Initial velocity = $v_{y1}$

Final velocity:

$$v_{y2} = v_{y1} - a t_1$$ $$= a t_1 - a t_1 = 0$$

So particle exits horizontally.

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Step 4: Net displacement

Second half displacement:

$$y_2 = v_{y1}t_1 - \frac{1}{2}a t_1^2$$

Substitute:

$$y_2 = a t_1^2 - \frac{1}{2}a t_1^2$$ $$= \frac{1}{2}a t_1^2$$

Total:

$$y = y_1 + y_2 = a t_1^2$$ $$= a \frac{L^2}{4u^2}$$

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🔴 PROBLEM 3

Electric Field + Gravity

Net acceleration:

$$a_{net} = \frac{qE}{m} - g$$

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(1) Straight line condition

$$a_{net} = 0$$ $$qE = mg$$

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(2) If $qE > mg$

Net upward acceleration ⇒ upward parabola.

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(3) If $qE < mg$

Net downward acceleration ⇒ downward parabola.

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🔴 PROBLEM 4

Time Minimization

Time:

$$t = \frac{L}{u\cos\theta}$$

To minimize time ⇒ maximize $\cos\theta$

Maximum at:

$$\theta = 0$$

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Deflection:

$$y = \frac{qEL^2}{2mu^2\cos^2\theta}$$

Minimize:

Take derivative wrt $\theta$

Minimum at:

$$\theta = 0$$

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🔴 PROBLEM 5

Velocity Doubles

Given:

$$v = 2u$$

Use:

$$v^2 = u^2 + 2ad$$ $$4u^2 = u^2 + 2ad$$ $$3u^2 = 2ad$$ $$a = \frac{3u^2}{2d}$$

But:

$$a = \frac{qE}{m}$$ $$E = \frac{3mu^2}{2qd}$$

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Time taken

$$v = u + at$$ $$2u = u + at$$ $$t = \frac{u}{a}$$

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Work done

$$W = qEd$$ $$= \frac{3}{2}mu^2$$

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Potential difference

$$V = Ed$$ $$= \frac{3mu^2}{2q}$$

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🔴 PROBLEM 6

Re-Entry Condition

A particle enters perpendicular to field and exits at angle:

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

After exiting, there is no field, so motion is straight line with:

Horizontal velocity:

$$v_x = u$$

Vertical velocity at exit:

$$v_y = \frac{qEL}{mu}$$

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Step 1: Equation of path after exit

Straight line:

$$y = L\tan\theta + (x-L)\tan\theta$$ $$y = x\tan\theta$$

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Step 2: Re-entry condition

Re-entry happens if trajectory intersects entrance boundary (x=0 region).

Slope must be large enough so that backward extension hits lower plate.

Geometric condition:

$$\tan\theta > \frac{d}{L}$$

Substitute:

$$\frac{qEL}{mu^2} > \frac{d}{L}$$ $$\boxed{\frac{qEL^2}{mu^2 d} > 1}$$

That is the re-entry condition.

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🔴 PROBLEM 7

Minimum Speed to Avoid Plate Collision

Maximum vertical deflection inside field:

$$y = \frac{qEL^2}{2mu^2}$$

To avoid collision:

$$y \le \frac{d}{2}$$ $$\frac{qEL^2}{2mu^2} \le \frac{d}{2}$$

Solve:

$$u^2 \ge \frac{qEL^2}{md}$$ $$\boxed{u_{min} = \sqrt{\frac{qEL^2}{md}}}$$

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🔴 PROBLEM 8

Moving Frame Analysis (Deep Concept)

Observer moves with speed $v$ horizontally.

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Step 1: Velocity transformation

$$u'_x = u - v$$

Vertical velocity unchanged.

Acceleration unchanged.

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Step 2: Equations in moving frame

$$x' = (u-v)t$$ $$y = \frac{1}{2}at^2$$

Eliminate $t$:

$$t = \frac{x'}{u-v}$$

Substitute:

$$y = \frac{a}{2(u-v)^2}x'^2$$

Still a parabola.

✔ Shape invariant under Galilean transformation.

This is a very high-level conceptual insight.

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🔴 PROBLEM 9

Same KE: Electron, Proton, Alpha

Given same KE:

$$\frac{1}{2}mv^2 = K$$ $$v^2 = \frac{2K}{m}$$

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Step 1: Deflection

$$y = \frac{qEL^2}{2m v^2}$$

Substitute $v^2$:

$$y = \frac{qEL^2}{2m \frac{2K}{m}}$$ $$y = \frac{qEL^2}{4K}$$

Mass cancels!

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Therefore:

Deflection depends only on charge.

Electron: $q=e$
Proton: $q=e$
Alpha: $q=2e$

Ratio:

$$y_e : y_p : y_\alpha = 1 : 1 : 2$$

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Step 2: Time of flight

$$t = \frac{L}{v}$$ $$t \propto \sqrt{m}$$ $$t_e : t_p : t_\alpha = \sqrt{m_e} : \sqrt{m_p} : \sqrt{4m_p}$$ $$= \sqrt{m_e} : \sqrt{m_p} : 2\sqrt{m_p}$$

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🔴 PROBLEM 10

Electric Field + Gravity Projectile

Net acceleration:

$$a = \frac{qE}{m} - g$$

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Condition for purely horizontal motion

At some time:

$$v_y = 0$$ $$u\sin\theta - at = 0$$ $$t = \frac{u\sin\theta}{a}$$

Must be real and positive:

$$a > 0$$ $$\frac{qE}{m} > g$$

If electric force exceeds weight, particle behaves like upward-accelerated projectile.

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🎯 Summary of Advanced Insights

Concept	Deep Result
Same KE particles	Deflection depends only on charge
Field reversal	Exit horizontal
Moving frame	Parabola remains parabola
Re-entry	Geometry inequality condition
Minimum speed	Inverse square relation