Question

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 2.0 cm, and the electric field within the capacitor has a magnitude of 2.3 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Answer 1

7.36×10⁻¹⁵ J

Step-by-step explanation:

From the question,

Then kinetic energy of the electron = electric potential × charge of the electron

Ek = q'V......................... Equation 1

Where Ek = kinetic Energy of the electron, q' = charge of an electron, V = Electric potential.

But,

V = E×d.................... Equation 2

Where E = Electric Field, d = distance of separation of the plates of the capacitor

Given: E = 2.3×10⁶ V/m, d = 2.0 cm = (2/100) m = 0.02 m.

Substitute into equation 2

V = 2.3×10⁶(0.02)

V = 4.6×10⁴ V.

Constant: q' = 1.6×10⁻¹⁹ C

Substitute into equation 1

Ek = 4.6×10⁴(1.6×10⁻¹⁹)

Ek = 7.36×10⁻¹⁵ J

Answer 2

Answer: 7.37×10^-15 J

Explanation: according to the work- energy theorem.

The work done in moving the electron from the negative plate to the positive equals it kinetic energy.

Mathematically, we have that

Kinetic energy = work done = qV

The distance traveled by the electron is simply the the distance between the plates of the capacitor (d) = 2cm = 0.02m.

Strength of electric field (E) = 2.3×10^6 v/m

We need to get the potential difference between the plates first, this is gotten by using the formulae below

V = Ed

Where V = potential difference =?

E = strength of electric field = 2.3×10^6 v/m

d = distance between plates = 0.02m

By substituting the parameters, we have that

V = 2.3×10^6 × 0.02

V = 0.046× 10^6

V = 4.6×10^4 v

But work done = qV

Where q = magnitude of electronic charge = 1.602×10^-19c

Work done = 1.602×10^-19 × 4.6×10^4

Work done = 7.37×10^-15 J