Question 877: Electrostats & Capacitance

Question

An electron of mass '$$m$$' and charge '$$q$$' is accelerated from rest in a uniform electric field of strength '$$E$$'. The velocity acquired by the electron when it travels a distance '$$L$$' is

Accepted Answer

The correct option for the velocity acquired by an electron when it travels a distance '$$L$$' in a uniform electric field of strength '$$E$$' is found by considering the work done by the electric field on the electron and equating it to the kinetic energy gained by the electron.
The work done by the electric field on the electron is given by the product of the force ($$F$$) exerted by the electric field on the electron, and the distance ($$L$$) over which this force acts. The force exerted by the electric field on a charge is given by $$F = qE$$, where $$q$$ is the charge of the electron, and $$E$$ is the strength of the electric field. Therefore, the work done ($$W$$) by the electric field can be expressed as:
$$W = F \cdot L = qEL$$
According to the principle of conservation of energy, the work done by the electric field is converted into kinetic energy ($$K$$) of the electron. The kinetic energy of an object with mass $$m$$ and velocity $$v$$ is given by $$K = \frac{1}{2}mv^2$$. Therefore, we equate the work done to the kinetic energy to find the velocity:
$$qEL = \frac{1}{2}mv^2$$
Solving this equation for $$v$$ to find the velocity of the electron after traversing the distance $$L$$ yields:
$$v = \sqrt{\frac{2qEL}{m}}$$
Hence, the correct answer is
Option C: $$\sqrt{\frac{2qEL}{m}}$$

Answer

$$\sqrt{\frac{2 \mathrm{qE}}{\mathrm{mL}}}$$

Answer

$$\sqrt{\frac{2 \mathrm{Em}}{\mathrm{qL}}}$$

Answer

$$\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}$$

Answer

$$	imes \sqrt{\frac{\mathrm{qE}}{\mathrm{mL}}}$$

Electrostats & Capacitance

An electron of mass '$m' and charge 'q' is accelerated from rest in a uniform electric field of strength 'E'. The velocity acquired by the electron when it travels a distance 'L$' is

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