<span>The speed of a wave is dependent upon the properties of the medium through which the wave is moving. An alteration in the properties of the medium will result in a change in the speed at which the wave moves through that medium.The speed of a wave is independent of its frequency and wavelength. A change in the frequency value will not result in a change in the speed value. Rather, changing frequency will result in a change in the wavelength in such a manner that the speed value turns out the same.</span>
Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charge density of 49.0 nC/m2. A proton is released from rest at the positive plate.
(a) Determine the magnitude of the electric field between the plates from the charge density.
b) Determine the potential difference between the plates.
Answer:
a
The Electric Field between the two plate is
b
The potential difference between the plate is V =
Explanation:
From the we are given that
The separation between the plate is
The surface charge density is
Generally Electric field between the plate is mathematically given as
Note that is the permitivity of free space and its value is
Now substituting values we have
Generally Potential difference between the plate is mathematically given as
Where E is the electric field which is
Substituting value we have
V =
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz