Respuesta :
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:
Given :
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 ( [tex]K_{A}[/tex]= 0) to the point B at distance l = 0.500m where its kinetic energy is ( [tex]K_{B}[/tex]= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.
Required :
(a) We are asked to find the electric potential VB
(b) We want to determine the magnitude and the direction of the electric field E.
Solution
(a) We are given the values for VA,[tex]K_{B}[/tex] 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:
[tex]K_{A} +U_{A} =K_{B} +U_{B}[/tex] .........................................(1)
Where [tex]K_{A}[/tex]= 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=[tex]K_{B}[/tex] +qVB (Divide by q)
VA=[tex]K_{B}[/tex] /q + VB (solve for VB)
VB=VA- [tex]K_{B}[/tex]/q .......................................(2)
We get the relation between VB, VA and [tex]K_{B}[/tex], now we can plug our values for VA, [tex]K_{B}[/tex] and q into equation (2) to get VB
VB=VA- [tex]K_{B}[/tex]/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 =[tex]\int\limits^1_0 {E} \, dl[/tex]...................................(a)
[tex]=E\int\limits^1_0 {} \, dl[/tex]
VA-VB=E[tex]l[/tex] (solve for E)
E= VA-VB/[tex]l[/tex]..................................(3)
Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E
E= VA-VB/[tex]l[/tex]
=-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
The electric potential at point B is -83.33 V, electric field is 226.7 V/m. and the electric field will be directed opposite to the negative charge.
Electric potential at point B
The electric potential at point B is calculated as follows;
K.E = qV
V = K.E/q
V = (5 x 10⁻⁷)/(-6 x 10⁻⁹)
V = - 83.33 V
Magnitude of the Electric field
The magnitude of the electric field is calculated as follows;
E = -Vba/d
E = - (Vb - Va)/d
E = - (-83.33 - 30)/0.5
E = 226.7 V/m
Direction of the electric field
The electric field will be directed opposite to the negative charge.
Learn more about electric potential and electric field here: https://brainly.com/question/24317284
