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Elan Coil [88]
3 years ago
8

A solid metallic sphere of radius a carries total charge Q. No other charges are nearby. The electric field just outside its sur

face is keQ/a2 radially outward. At this close point, the uniformly charged surface of the sphere looks exactly like a uniform flat sheet of charge. What is the electric field strength inside the metallic sphere at r = a/2? Use σ = Q/4 π a2.
Physics
1 answer:
NNADVOKAT [17]3 years ago
3 0
Please see below solution:

The electric field just outside its surface is = 2 , where is a radius of sphere, is a charge of sphere, = 1 4ε0 . The electric field is = 1 4ε0 2 = 1 ε0 ∙ = σ ε0 , where = 4 2 is an area of sphere, σ = Q S is a surface charge density. Answer: . 
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A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

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