Answer:
a) 80 V
b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge
Explanation:
<u>Given
:</u>
We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero (  = 0) to the point B at distance l = 0.500m where its kinetic energy is
(
= 0) to the point B at distance l = 0.500m where its kinetic energy is
(   = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = +
30.0 v.
= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = +
30.0 v.
<u>Required
:</u>
<em>(a)	We are asked to find the electric potential VB
</em>
<em>(b)	We want to determine the magnitude and the direction of the electric field E.
</em>
<u>
Solution
</u>
(a)	We are given the values for VA, and q so we want to find a relationship between these three parameters and VB to get the value of VB.
 and q so we want to find a relationship between these three parameters and VB to get the value of VB.
As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:
                                     .........................................(1)
 .........................................(1)                                           
Where  = 0 and the potential energy U of the charge is given by U = q V
= 0 and the potential energy U of the charge is given by U = q V
where V is the electric potential.  So, equation (1) will be in the form
:
                                   0+qVA= +qVB                      (Divide by q)
 +qVB                      (Divide by q)
                                          VA= /q + VB                  (solve for VB)
 /q + VB                  (solve for VB)
                                          VB=VA-  /q .......................................(2)
/q .......................................(2)
We get the relation between VB, VA and  , now we can plug our values for VA,
, now we can plug our values for VA,  and q into equation (2) to get VB
 and q into equation (2) to get VB
                                          VB=VA-  /q
/q
                                               =30V-(5.00 x 10^-7J)/(-6.00 x I0^-9) 
                                               =80 V
(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points
                                    VA-VB = ...................................(a)
...................................(a)
                                                
                                    VA-VB=E (solve for E)
                      (solve for E)
                                             E= VA-VB/ ..................................(3)
..................................(3)
Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E
                                             E= VA-VB/
                                               =-100 N/C
The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge