The Question is from neet-physics electrostatic-charges-and-fields , and it was created by the author P.NARASIMHA

Question

A charged particle is projected with a velocity $v$ at an angle $	heta$ to a uniform electric field $E$.  If the particle's velocity becomes perpendicular to the field after time $t$,  then what is the initial velocity component perpendicular to the field?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Projectile motion under a uniform electric field
	**Related Topics**: Kinematics, electric force, vectors
	**Prerequisites**: Understanding of vector components, projectile motion, and the effect of a uniform electric field on a charged particle.
## SOLUTION PATHWAY
**Given Information**:
		Initial velocity = $v$ at angle $	heta$ to the electric field $E$
		Time $t$ when velocity becomes perpendicular to the field
		Charge of particle = $q$ (implicitly assumed)
		Mass of particle = $m$ (implicitly assumed)
**Method & Steps**:
	1. Resolve the initial velocity into components parallel and perpendicular to the field.
		Initial velocity parallel to the field: $v_{\parallel} = v \cos	heta$
		Initial velocity perpendicular to the field: $v_{\perp} = v \sin	heta$
	2. Analyze the effect of the electric field on the velocity components.
		The electric field only affects the velocity component parallel to the field. The acceleration due to the electric field is given by $a = \frac{qE}{m}$
		The velocity parallel to the field at time t is $v_{\parallel}(t) = v \cos	heta - \frac{qEt}{m}$
	3. Determine the condition when the final velocity is perpendicular to the field. This happens when the parallel component of the velocity becomes zero.
		$v_{\parallel}(t) = 0$
		$v \cos	heta - \frac{qEt}{m} = 0$
		This equation gives the relationship between the initial parallel velocity, electric field, charge, mass, and time when the final velocity is perpendicular to the field.
	4. The initial velocity component perpendicular to the field remains unchanged because the electric field does not exert a force in that direction. Therefore, the initial velocity component perpendicular to the field is $v \sin	heta$.
		Result: $v\sin	heta$
		Verification: The condition $v_{\parallel}(t) = 0$ ensures that the final velocity is perpendicular to the electric field. The perpendicular component remains unaffected.
## OPTION ANALYSIS
[A] :x: $v\cos	heta$: This is the initial velocity component parallel to the field.
[B] ✓: $v\sin	heta$: This is the initial velocity component perpendicular to the field, which remains constant.
[C] :x: $qEt/m$: This is the change in velocity parallel to the field after time t.
[D] :x: $v\cos	heta - qEt/m$: This represents the velocity component parallel to the field at time t, which becomes zero when the final velocity is perpendicular to the field.
## LEARNING AIDS
	**Common Mistakes**:
		1.Confusing parallel and perpendicular components: Carefully resolve the initial velocity into its components.
		2.Assuming the electric field affects the perpendicular component: The electric field only accelerates the particle along its own direction.
	**Quick Tips**:
		1.Draw a diagram to visualize the velocity components and the electric field.
		2.Remember that the electric field only affects the parallel component of velocity.

P.NARASIMHA · Answer

$v\cos	heta$

P.NARASIMHA · Answer

The correct Option is "$v\sin	heta$"

P.NARASIMHA · Answer

$qEt/m$

P.NARASIMHA · Answer

$v\cos	heta - qEt/m$

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