contestada

Moving 2.0 coulombs of charge at constant speed a distance of 6.0 meters from point a to pointb within a uniform electric field requires a 5.0-newton force. what is the electric potentialdifference between points a and b?

Respuesta :

skyluke89
The force produced on a particle of charge q by an electric field of intensity E is
[tex]F=qE[/tex]
in our problem, the force is F=5.0 N while the charge is q=2.0 C, so we can find the intensity of the electric field:
[tex]E= \frac{F}{q}= \frac{5.0 N}{2.0 C}=2.5 N/C [/tex]

The relationship between electric field intensity and potential difference [tex]\Delta V[/tex] between two points A and B is
[tex]\Delta V = V_B - V_A = - E d[/tex]
where d is the distance between the two points. By using d=6.0 m, we find
[tex]\Delta V = V_B-V_A = -(2.5 N/C)(6.0 m)=-15 V [/tex]
where the negative sign means that the initial point, VA, is at higher potential than the final point VB.
okpalawalter8

The electric potential difference between  a and b is mathematically given as

dV=-15 V

What is the electric potential difference between points a and b?

Question Parameters:

Moving 2.0 coulombs of charge at constant speed a distance of 6.0 meters

a uniform electric field requires a 5.0-newton force.

Generally, the equation for the  Force is mathematically given as

F=qE

Thereofre

E=F/q

E=5/2

E=2.5

Therefore

dV = V_B - V_A = - E d

dV = V_B-V_A

dV= -(2.5 N/C)(6.0 m)

dV=-15 V

Read more about Voltage

https://brainly.com/question/14883923

Otras preguntas