Question

A pair of charged conducting plates produces a uniform field of 12,000 N/C directed to the right, between the plates. The separation of the plates is 40 mm. In the figure, an electron is projected from plate A, directly toward plate B, with an initial velocity of 2.0 x 10^7 m/s. The velocity of the electron as it strikes plate B is closest to:

(a) 1.8 × 107 m/s

(b) 1.5 × 107 m/s

(c) 2.1 × 107 m/s

(d) 2.4 × 107 m/s

(e) 1.2 × 107 m/s

Answer 1

Concepts and reason

The concepts used to solve this problem are electric field, electric force, and Newton’s second law.

Initially compare the electric force exerted between the pair of charged conducting plates to push the electron from plate A to plate B and with Newton second law of motion to find the acceleration of the charged particle electron.

The force of attractive or repulsive interaction between the pair of charged conducting plates is called the electric force and this force acts on the conducting plates to produce an electrons tend to move from plate A to plate B with initial velocity.

Fundamentals

The electric force acting on the pair of charged conducting plates produce an negative charge of an electron projected from plate A to plate B in a uniform electric field directed towards right side between the plates.

The expression for the electric force acting on the charge of an electron produces a uniform electric field is,

$F=qE $

Here, is the electric force acting, is the charge of an electron, and is the electric field acting between the plates.

The electric field is the region around the space where the charged particle influences its force on other charged particle. Here, the pair of charged conducting plates produces a uniform electric field.

According to Newton’s second law, the force acting on the electron moving from plate A to plate B is,

$F=ma $

Here, is the force acting on the electron projected in between two plates, is the mass of the electron and is the acceleration of the electron moving in a uniform electric field.

According to kinematics equation, the velocity of the electron strikes on the plate B is,

Here, is the velocity of the electron strikes on the plate B, is the initial velocity of the plate A, is the acceleration of an electron moving from plate A to plate B, and is the separation distance between the plates.

The expression for the electric force acting on the charged particle of an electron produces a uniform electric field is,

$F=qE $ …. (1)

The expression for the force acting on the electron moving from plate A to plate B is,

$F=ma $ …. (2)

Compare the expression (1) and (2) to calculate the acceleration of the electron moving in a uniform electric field.

Substitute for , for , and for to find acceleration of the electron.

The velocity of the electron strikes on the plate B is,

Rewrite the expression in terms of .

Substitute for , for , and for to find the velocity of an electron strikes on the plate B.

Ans:

The velocity of the electron strikes on the plate B is.