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Answer:
b) bend away from the normal
Explanation:
According to snell's law , if i be the angle of incidence and r be the angle of refraction
sin i / sin r = 1.33 / 2.42
sin i / sin r = .55
Hence sin r > sin i
r > i
In other words angle of refraction will be more than angle of incidence . So, the ray will bend away from the normal .
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
a E =![1.685*10^3 N/C](https://tex.z-dn.net/?f=1.685%2A10%5E3%20N%2FC)
b E =![36.69*10^3 N/C](https://tex.z-dn.net/?f=36.69%2A10%5E3%20N%2FC)
c E = 0 N/C
d ![V = 6.7*10^3 V](https://tex.z-dn.net/?f=V%20%3D%206.7%2A10%5E3%20V)
e ![V = 26.79*10^3V](https://tex.z-dn.net/?f=V%20%3D%2026.79%2A10%5E3V)
f V = ![34.67 *10^3 V](https://tex.z-dn.net/?f=34.67%20%2A10%5E3%20V)
g ![V= 44.95*10^3 V](https://tex.z-dn.net/?f=V%3D%2044.95%2A10%5E3%20V)
h ![V= 44.95*10^3 V](https://tex.z-dn.net/?f=V%3D%2044.95%2A10%5E3%20V)
i ![V= 44.95*10^3 V](https://tex.z-dn.net/?f=V%3D%2044.95%2A10%5E3%20V)
Explanation:
From the question we are given that
The first charge ![q_1 = 2.00 \mu C = 2.00*10^{-6} C](https://tex.z-dn.net/?f=q_1%20%3D%202.00%20%5Cmu%20C%20%3D%202.00%2A10%5E%7B-6%7D%20C)
The second charge ![q_2 =1.00 \muC = 1.00*10^{-6}](https://tex.z-dn.net/?f=q_2%20%3D1.00%20%5CmuC%20%3D%201.00%2A10%5E%7B-6%7D)
The first radius ![R_1 = 0.500m](https://tex.z-dn.net/?f=R_1%20%3D%200.500m)
The second radius ![R_2 = 1.00m](https://tex.z-dn.net/?f=R_2%20%3D%201.00m)
![Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}](https://tex.z-dn.net/?f=Generally%20%5C%20Electric%20%5C%20field%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_1%2B%5C%20q_2%7D%7Br%5E2%7D)
And ![Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]](https://tex.z-dn.net/?f=Potential%20%5C%20Difference%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%20%20%5B%5Cfrac%7Bq_1%20%7D%7Br%7D%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%5D)
The objective is to obtain the the magnitude of electric for different cases
And the potential difference for other cases
Considering a
r = 4.00 m
![E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%28%282%2B1%29%2A10%5E%7B-6%7D%29%2A8.99%2A10%5E9%7D%7B16%7D)
![= 1.685*10^3 N/C](https://tex.z-dn.net/?f=%3D%201.685%2A10%5E3%20N%2FC)
Considering b
![r = 0.700 m \ , R_2 > r > R_1](https://tex.z-dn.net/?f=r%20%3D%200.700%20m%20%5C%20%2C%20R_2%20%3E%20r%20%3E%20R_1)
This implies that the electric field would be
![E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%5Cfrac%7Bq_1%7D%7Br%5E2%7D)
This because it the electric filed of the charge which is below it in distance that it would feel
![E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}](https://tex.z-dn.net/?f=E%20%3D%208%2A99%2A10%5E9%20%20%5Cfrac%7B2%2A10%5E%7B-6%7D%7D%7B0.4900%7D)
= ![36.69*10^3 N/C](https://tex.z-dn.net/?f=36.69%2A10%5E3%20N%2FC)
Considering c
r = 0.200 m
=> ![r](https://tex.z-dn.net/?f=r%3CR_1%3CR_2)
The electric field = 0
This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field
Considering d
r = 4.00 m
=> ![r > R_1 >r>R_2](https://tex.z-dn.net/?f=r%20%3E%20R_1%20%3Er%3ER_2)
Now the potential difference is
![V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V](https://tex.z-dn.net/?f=V%20%3D%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5Cfrac%7Bq_1%20%2B%20%5C%20q_2%7D%7Br%7D%20%3D%208.99%2A10%5E9%20%2A%20%5Cfrac%7B3%2A10%5E%7B-6%7D%7D%7B4%7D%20%3D%206.7%2A10%5E3%20V)
This so because the distance between the charge we are considering is further than the two charges given
Considering e
r = 1.00 m ![R_2 = r > R_1](https://tex.z-dn.net/?f=R_2%20%3D%20r%20%3E%20R_1)
![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5Cfrac%7B1.00%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2026.79%20%2A10%5E3%20V)
Considering f
![r = 0.700 m \ , R_2 > r > R_1](https://tex.z-dn.net/?f=r%20%3D%200.700%20m%20%5C%20%2C%20R_2%20%3E%20r%20%3E%20R_1)
![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.700%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2034.67%20%2A10%5E3%20V)
Considering g
![r =0.500\m , R_1 >r =R_1](https://tex.z-dn.net/?f=r%20%3D0.500%5Cm%20%2C%20R_1%20%3Er%20%3DR_1)
![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)
Considering h
![r =0.200\m , R_1 >R_1>r](https://tex.z-dn.net/?f=r%20%3D0.200%5Cm%20%2C%20R_1%20%3ER_1%3Er)
![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7BR_1%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)
Considering i
![r =0\ m \ , R_1 >R_1>r](https://tex.z-dn.net/?f=r%20%3D0%5C%20m%20%5C%20%2C%20R_1%20%3ER_1%3Er)
![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7BR_1%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)