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RUDIKE [14]
3 years ago
5

A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

l to the ground, and a vertical uniform electric field of strength 260N/C passes upward through the tetrahedron.
What is the electric flux through the base?







What is the electric flux through each of the three sides?
Physics
1 answer:
Jet001 [13]3 years ago
6 0

Answer:

-7.03645640575 Nm²/C

2.34548546858 Nm²/C

Explanation:

S = Surface area

Electric flux through the base is given by

\phi_b=-\int Eds\\\Rightarrow \phi_b=-ES\\\Rightarrow \phi_b=-260\dfrac{\sqrt{3}}{4}0.25^2\\\Rightarrow \phi_b=-7.03645640575\ Nm^2/C

The electric flux through the base is -7.03645640575 Nm²/C

Inside the tetrahedron there is no charge so total flux will be zero

\phi_b+3\phi_s=0\\\Rightarrow \phi_s=-\dfrac{\phi_b}{3}\\\Rightarrow \phi_s=-\dfrac{-7.03645640575}{3}\\\Rightarrow \phi_s=2.34548546858\ Nm^2/C

The electric flux through the sides is 2.34548546858 Nm²/C

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m = 43.2 kg

Explanation:

volume of sphere = (4/3)pi(r)^3

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3 years ago
Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


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