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?

7.36×10⁻¹⁵ J

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

no. dot product   =0 when vecs at right angles

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force