The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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Answer:
<em>Part A</em><em>:</em>
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
<em>Part B</em><em>:</em>
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
<em>Part C</em><em>:</em>
a) If the distance to the screen is decreased the fringe spacing will decrease.
<em>Part D</em><em>:</em>
The dot in the center of fringe E is 
farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with
- m is the order number.
is the wavelength of the monochromatic light.- L is the distance between the screen and the two slits.
- d is the distance between the slits.
- Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
- Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
- Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
- Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be
We simply replace the values in that equation :
The dot in the center of fringe E is 
farther from the left slit than from the right slit.
Answer:
0.80 m
Explanation:
elastic potential energy formula
elastic potential energy = 0.5 × spring constant × (extension) 2
Incomplete question. Missing part:
"it took him 10 seconds to move the couch. What was his power?"
Answer:
50 W
Explanation:
The power developed is given by:
where
W is the work done
t is the time taken
In this problem,
W = 500 J
t = 10.0 s
Substituting into the equation, we find
