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Question

(A) The change in kinetic energy $(\Delta K)$ is equal to the total work done by all forces acting on the particle.
Since the magnetic force $\vec{F}

Vedclass · Accepted Answer

$\frac{qE_0}{\lambda}[1 - e^{-\lambda L}]$

Vedclass · Answer

(A) The change in kinetic energy $(\Delta K)$ is equal to the total work done by all forces acting on the particle.
Since the magnetic force $\vec{F}_m = q(\vec{v} 	imes \vec{B})$ is always perpendicular to the velocity $\vec{v}$, it does no work on the particle.
Therefore, the work is done only by the electric force $\vec{F}_e = q\vec{E} = qE_0 e^{-\lambda x}\hat{i}$.
The work done $W$ as the particle moves from $x = 0$ to $x = L$ is given by:
$W = \int_{0}^{L} F_x dx = \int_{0}^{L} q E_0 e^{-\lambda x} dx$
$W = q E_0 \left[ \frac{e^{-\lambda x}}{-\lambda} ight]_{0}^{L}$
$W = \frac{q E_0}{-\lambda} (e^{-\lambda L} - e^0) = \frac{q E_0}{\lambda} (1 - e^{-\lambda L})$.
Thus, the change in kinetic energy is $\frac{q E_0}{\lambda} (1 - e^{-\lambda L})$.

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