Question

A uniform electric field of 200 N/C is in the x-direction. A point charge of
3μC
is released from rest at the origin. What is the kinetic energy of the charge when it is at x = 4 m?
2.4×10−2J
1.6×10−2J
3.6×10−2J
4.8×10−2J

Answer 1

Answer on Question #55896, Physics / Electromagnetism

Task: A uniform electric field of $200\,\mathrm{N/C}$ is in the x-direction. A point charge of $3\,\mu\mathrm{C}$ is released from rest at the origin. What is the kinetic energy of the charge when it is at $x = 4\,\mathrm{m}$?

$$\begin{array}{l} 2.4 \times 10^{-2}\,\mathrm{J} \\ 1.6 \times 10^{-2}\,\mathrm{J} \\ 3.6 \times 10^{-2}\,\mathrm{J} \\ 4.8 \times 10^{-2}\,\mathrm{J} \\ \end{array}$$

Solution:

Force on charge: $F = E\,q$

This force is constant and will increase charge's velocity with constant acceleration.

By the Second Newton's law: $F = m\,a \rightarrow a = F/m = E\,q/m$

Thus, time needed to reach $x$: $x = \frac{at^2}{2} \Rightarrow t = \sqrt{\frac{2x}{a}} = \sqrt{\frac{2x\,m}{Eq}}$

Charge velocity at $x$: $v = at = \frac{E\,q}{m} \sqrt{\frac{2mx}{Eq}} = \sqrt{\frac{2x\,E\,q}{m}}$

And kinetic energy: $K = \frac{mv^2}{2} = \frac{m}{2} \cdot \frac{2x\,E\,q}{m} = E\,x\,q = 200\,\mathrm{N/C} \cdot 3 \cdot 10^{-6}\,\mathrm{C} \cdot 4\,m = 2.4 \cdot 10^{-3}\,\mathrm{J}$

Answer: kinetic energy of the charge $K = 2.4 \cdot 10^{-3}\,\mathrm{J}$

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