Charged Particle in Electric Field PYQ | JEE Main Physics 2025

❓ Question

Two charged particles have charges $q_1,q_2$ and masses $m_1,m_2$.

Initially,

$$v_i=u$$

for both particles, and they move for the same time $T$ in a uniform electric field $E$.

For particle 1,

• Final speed $=\dfrac{u}{4}$
• Angle between initial and final velocity $=60^\circ$
  

For particle 2,

• Final speed $=\ ?$
  
• Angle between initial and final velocity $=90^\circ$
  

The electric field $E$ makes an angle with the initial velocity $u$. Find the final speed of particle 2.

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✍️ Solution

Let the electric field make an angle $\theta$ with the initial velocity.

The acceleration due to the electric field is

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

Only the component of acceleration along the initial velocity changes the horizontal component.

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Step 1: For Particle 1

Horizontal component of final velocity:

$$V_{x1}=u-\frac{qE_xT}{m_1}.$$

Since the final speed is

$$V_1=\frac{u}{4},$$

and the direction changes by $60^\circ$,

$$V_{x1}=V_1\cos60^\circ =\frac{u}{4}\cdot\frac12 =\frac{u}{8}.$$

Therefore,

$$\boxed{\frac{7u}{8}=\frac{qE_xT}{m_1}} \qquad (1)$$

Vertical component:

$$V_{y1}=\frac{qE_yT}{m_1}.$$

Also,

$$V_{y1}=V_1\sin60^\circ =\frac{u}{4}\cdot\frac{\sqrt3}{2} =\frac{u\sqrt3}{8}.$$

Hence,

$$\boxed{\frac{u\sqrt3}{8}=\frac{qE_yT}{m_1}} \qquad (2)$$

Dividing (2) by (1),

$$\frac{E_y}{E_x} = \frac{\sqrt3}{7}.$$

Thus,

$$\tan\theta=\frac{E_y}{E_x} =\frac1{\sqrt3},$$

so

$$\boxed{\theta=30^\circ.}$$

Therefore, the angle between $E$ and the initial velocity is

$$90^\circ-30^\circ=60^\circ.$$

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Step 2: For Particle 2

The final velocity is perpendicular to the initial velocity.

Hence,

$$V_{x2}=0.$$

Therefore,

$$u-\frac{qE_xT}{m_2}=0,$$

or

$$\boxed{u=\frac{qE_xT}{m_2}} \qquad (3)$$

Vertical component:

$$V_{y2}=\frac{qE_yT}{m_2}.$$

Using (3),

$$V_{y2} = u\cdot\frac{E_y}{E_x}.$$

Since

$$\frac{E_y}{E_x}=\frac1{\sqrt3},$$

we obtain

$$V_{y2}=\frac{u}{\sqrt3}.$$

As the horizontal component is zero,

$$V_2=V_{y2}.$$

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✅ Final Answer

$$\boxed{{\displaystyle \frac{u}{\sqrt{3}}}}$$