Question

Two 2.4 cm -diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.8×10^5 V/m.

Part A:
What is the voltage across the capacitor?
V= 960 V is correct

Part B:
How much charge is on each disk?

q1, q2= ?? C

*Definitely not sure how to do this one.

Part C:
An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0×10^7 m/s . What was the electron's speed as it left the negative plate?

viniticial= ?? m/s

*I already tried 6.2 x 1013 but that was not right.

Answer 1

Concepts and reason

The required concepts to solve the problem are capacitance of a capacitor, potential difference between the plates of a capacitor and the law of conservation of energy.

First, using the relation between the potential difference, electric field and the separation distance between the plates, find the voltage across the capacitor. Then, from the equations of capacitance, find the charge on each disk. Finally, using law of conservation of energy, the electron’s speed as it left the negative plate can be found.

Fundamentals

A parallel plate capacitor is a capacitor formed by two metal plates separated by a distance. The capacitance of a parallel plate capacitor is,

Here, is the permittivity of free space, is the area of cross-section of plates and is the plate separation distance.

The capacitance is the ratio of the charge and the potential difference. The relation between charge and potential for the capacitance of the capacitor is,

Here, is the charge in the capacitor and is the potential difference.

The voltage across the capacitor is the product of the electric field and the separation. Distance then the equation becomes,

Here, is the electric field.

According to the law of conservation of energy, the initial energy is equal to the final energy.

Here, is the initial energy andis the final energy.

(A)

The equation for voltage across the capacitor is,

Substitute for and for .

(B)

The area of the plate is,

…… (1)

Here, is the radius.

The radius of the disk is,

Here, is the diameter of the disk.

Replace for r in the equation (1).

…… (2)

The expression for capacitance of parallel plate capacitor is,

Substitute equation (2) in above equation.

…… (3)

The capacitance of a capacitor is,

…… (4)

Equating the equations (3) and (4) of the capacitance ,to find the charge of the capacitor,

Substitute for , for , for and for .

(C)

According to the law of conservation of energy,

The initial energy is the sum of the initial kinetic energy and potential energy and final energy is the sum of the final kinetic energy and potential energy.

Substitute for and for .

Here, is the initial kinetic energy, is the final kinetic energy, is the initial potential energy andis the final potential energy.

Substitute for , for, for and for .

Here, is the mass of the electron, is the electron charge, is the velocity of electron when it leaves the negative plate and is the velocity of electron as it reaches the positive plate.

Thus, the speed of the electron as it leaves the negative plate is,

Substitute for , for , forandfor.

Ans: Part A

The voltage across the capacitor is.

Part B

The charge on each disk is.

Part C

The speed of the electron as it leaves the negative plate is .