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

Question

An electron enters a uniform electric field $E$ with an initial velocity $v_0$ perpendicular to the field.  What is the trajectory of the electron?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Motion of a charged particle in a uniform electric field
 **Related Topics**: Projectile motion, kinematics, electric force
 **Prerequisites**: Understanding of electric fields, forces, Newton's laws of motion, and basic kinematics.
## SOLUTION PATHWAY
**Given Information**:
  * Uniform electric field $E$
  * Initial velocity of electron $v_0$ perpendicular to the field
  * Mass of electron $m$ (implicitly considered)
  * Charge of electron $q$ (implicitly considered, negative)
**Method & Steps**:
 1. Analyze the forces acting on the electron.
  Reasoning: The only force acting on the electron is the electric force due to the uniform electric field, given by $F = qE$.  Since the field is uniform, this force is constant.
  Formula/Rule: $F = qE$
 2. Determine the acceleration of the electron.
  Working: Using Newton's second law, $F = ma$, the acceleration of the electron is $a = \frac{F}{m} = \frac{qE}{m}$. This acceleration is constant and in the direction of the electric field (opposite to the direction of the electron's initial velocity since the charge is negative).
  Key Point: The acceleration is constant and in one direction only (parallel to the electric field).
 3. Describe the trajectory based on the constant acceleration and initial velocity.
  Result: Since the electron has an initial velocity perpendicular to the acceleration, it will follow a parabolic trajectory. This is analogous to projectile motion under gravity, where gravity provides a constant downward acceleration.
  Verification: The equation of motion in the direction of the field is $y = v_{0y}t + \frac{1}{2}at^2 = \frac{1}{2}(\frac{qE}{m})t^2$ (assuming $v_{0y} = 0$). In the perpendicular direction, $x = v_{0x}t$. Eliminating $t$, we get a parabolic equation relating $x$ and $y$.
 ## OPTION ANALYSIS
A :x: Circular motion requires a centripetal force always perpendicular to the velocity. The electric force here is constant and in one direction.
B :x: A straight line trajectory would only occur if there were no net force on the electron.
C ✓: Parabolic trajectory is the correct result due to constant acceleration in one direction and constant velocity in the perpendicular direction.
D :x: Helical motion requires a force component parallel to the velocity and a force component perpendicular to the velocity. This is not the case here.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Assuming the trajectory is circular due to the presence of a force:  Remember the direction of the force is crucial. A circular trajectory requires a force constantly perpendicular to the velocity.
  2.Ignoring the direction of the electric force relative to the initial velocity: The sign of the charge is critical in determining the direction of the acceleration.
 **Quick Tips**:
  1.Relate this problem to projectile motion under gravity. The mathematical description is very similar.
  2.Remember that a constant force results in constant acceleration, leading to parabolic motion when the initial velocity is perpendicular to the force.

SUBHANI · Answer

Circular

SUBHANI · Answer

Straight line

SUBHANI · Answer

The correct Option is "Parabolic"

SUBHANI · Answer

Helical

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