Answer:
The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m
Explanation:
Data
F = 2 N
d = ?
q = 1.6 x 10 ⁻¹⁹ C
k = 8.987 Nm²/C²
Formula
Solve for r
Substitution
Simplification
r =
r =
Result
r = 1.07 x 10⁻¹⁴ m
Answer:
1. 24375 N/C
2. 2925 V
Explanation:
d = 12 cm = 0.12 m
F = 3.9 x 10^-15 N
q = 1.6 x 10^-19 C
1. The relation between the electric field and the charge is given by
F = q E
So,
E = 24375 N/C
2. The potential difference and the electric field is related by the given relation.
V = E x d
where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.
By substituting the values, we get
V = 24375 x 0.12 = 2925 Volt
To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.
Mathematically the expression is given as
Where,
= Initial Intensity
I = Final Intensity after pass through the polarizer
= Angle between the polarizer and the light
Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as
Using the Malus Law we have,
Angle with respect to maximum is
Answer:
1 bright fringe every 33 cm.
Explanation:
The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:
where
m is the order of the maximum
is the wavelength of the light
D is the distance of the screen
d is the separation between two adjacent slit
Here we have:
is the wavelength of the light
D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)
is the number of lines per mm, so the spacing between two lines is
Therefore, substituting m = 1, we find:
So, on the distant screen, there is 1 bright fringe every 33 cm.