Question

A particle of mass \(m\) and charge \(q\) starts from rest and moves in an electric field \(\vec{E} = E_0 \hat{i}\). After travelling a distance \(x\) in the field along \(x\)-axis, the kinetic energy of the particle will be :

• A. \(q E_0 x^2\)
• B. \(q E_0 x\)
• C. \(q^2 E_0 x\)
• D. \(q^2 E_0^2 x^2\)

Hint:

For conservative fields like a uniform electric field, the kinetic energy gained starting from rest can be directly computed using \(K = q \cdot \Delta V\), where \(\Delta V = E \cdot x\). Hence, \(K = qE_0x\).

Answer 1

The Correct Option is B

Concept: When a charged particle is placed in an external electric field, it experiences an electrostatic force. This force accelerates the particle, thereby doing work on it. According to the Work-Energy Theorem, the net work done by all the forces acting on a particle is equal to the change in its kinetic energy.
• Electrostatic Force: The force \(\vec{F}\) acting on a charge \(q\) in an electric field \(\vec{E}\) is given by: \[ \vec{F} = q\vec{E} \]
• Work-Energy Theorem: The work done \(W\) by the force over a displacement \(\vec{d}\) is given by: \[ W = \int \vec{F} \cdot d\vec{r} = \Delta K = K_f - K_i \]

Step 1: Finding the force acting on the particle.
The given uniform electric field acts completely along the positive \(x\)-axis: \[ \vec{E} = E_0 \hat{i} \] The electrostatic force experienced by the charge \(q\) is: \[ \vec{F} = q\vec{E} = q(E_0 \hat{i}) = qE_0 \hat{i} \] Since the field is uniform, the magnitude of this force is a constant value: \[ F = qE_0 \]

Step 2: Calculating the work done and kinetic energy.
The particle travels a displacement of distance \(x\) along the direction of the force (positive \(x\)-axis). Thus, the work done by the electric field on the particle is: \[ W = F \cdot x = (qE_0) \cdot x = qE_0x \] Using the Work-Energy Theorem, the work done is equal to the change in kinetic energy: \[ W = K_f - K_i \] Since the particle starts from rest, its initial velocity is zero, which means its initial kinetic energy is zero (\(K_i = 0\)): \[ qE_0x = K_f - 0 \quad \Rightarrow \quad K_f = qE_0x \] Thus, the final kinetic energy of the particle after moving a distance \(x\) is \(qE_0x\).