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)
displ = velocity x time
25 x 3.2 = 75+5 km north.
A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.
So, the hang up time, or the time the object stayed on air is calculated using this equation:
a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds
When you touch an object and heat flows OUT of it, INTO your finger, you say the object feels hot.
When you touch an object and heat flows INTO it, OUT of your finger, you say the object feels cold.
If the object has the same temperature as your finger ... <em>around the mid-90s</em> ... then no heat flows in or out of your finger when you touch the object, and the object doesn't feel hot or cold.