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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 =
b E =
c E = 0 N/C
d 
e 
f V = 
g 
h 
i 
Explanation:
From the question we are given that
The first charge 
The second charge 
The first radius 
The second radius 

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


Considering b

This implies that the electric field would be

This because it the electric filed of the charge which is below it in distance that it would feel

= 
Considering c
r = 0.200 m
=> 
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
=> 
Now the potential difference is

This so because the distance between the charge we are considering is further than the two charges given
Considering e
r = 1.00 m 
![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

![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

![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

![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

![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)