Answer:
A) i) E =α/ [2πrL(εo)]
ii) E=0
iii) E = α/(πrεo)
The graph between E and r for the 3 cases is attached to this answer ;
B) i) charge on the inner surface per unit length = - α
ii) charge per unit length on the outer surface = 2α
Explanation:
A) i) For r < a, the charge is in the cavity and takes a shape of the cylinder. Thus, applying gauss law;
EA = Q(cavity) / εo
Now, Qcavity = αL
So, E(2πrL) = αL/εo
Making E the subject of the formula, we have;
E =α/ [2πrL(εo)]
ii) For a < r < b; since the distance will be in the bulk of the conductor, therefore, inside the conductor, the electric field will be zero.
So, E=0
iii) For r > b; the total enclosed charge in the system is the difference between the net charge and the charge in the inner surface of the cylinder.
Thus, Qencl = Qnet - Qinner
Qinner will be the negative of Qnet because it should be in the opposite charge of the cavity in order for the electric field to be zero. Thus;
Qencl = αL - (-αL) = 2αL
Thus, applying gauss law;
EA = Qencl / εo
Thus, E = Qencl / Aεo
E = 2αL/Aεo
Since A = 2πrL,
E = 2αL/2πrLεo = α/(πrεo)
B) i) The charge on the cavity wall must be the opposite of the point charge. Therefore, the charge per unit length in the inner surface of the tube will be - α
ii)Net charge per length for tube is +α and there is a charge of - α on the inner surface. Thus charge per unit length on the outer surface will be = +α - (- α) = 2α
Explanation :
It is given that,
BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :
We know that, 1 kilocalorie = 4184 joules
So,
J/sec is nothing but watts.
So,
and
So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.
Answer:
See explanation
Explanation:
Solution:-
- Here we will assume that the grating has the line density ( N ) defined by the number of lines per mm.
- The angle that each fringe forms on the screen is defined by ( θ ).
- The order of bright/dark spot is defined by an integer ( n )
- The wavelength of the incident light is ( λ )
- Here we will use the relation given for diffraction grating by the Young's Experiment as follows:
- The above given formulation is for constructive interference.
- We will inspect the effect of increasing the distance between the screen and the grating. Consider the length ( L ) from the center of the grating to the center of the screen. The distance ( yn ) will denote the distance between each fringe in vertical direction on the screen.
- For small angles ( θ ) we can make an approximation of sin ( θ ) ≈ tan ( θ ). Where,
sin ( θ ) ≈ tan ( θ ) = [ yn / L ]
- Substitute the above approximation in the given relation of diffraction gratings as follows:
- To double the distance between the screen and grating we will use the above relation with ( 2L ):
yn ∝ L
Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread! This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern. The spread also reduces the intensity/contrast between the bright and dark fringes because the distance travelled by each ray of light has increased. The intensity is inversely proportional to the square of distance travelled.
- Similarly, the line density of the grating ( N ) was doubled. Then,
yn ∝ N
Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread!This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern.