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)
 
        
             
        
        
        
Answer:
I don't know but I will try:
because the red color from the density column.