Uniform electric field. Find the potential difference

[Bearbull24.5]

Homework Statement

Point A (2,3)m and point B (5,7)m are in a region where electric field is uniform and given by E=(4i+3j)N/C. What is potential diff Va-Vb?

Homework Equations

Not a clue

The Attempt at a Solution

I tried converting the (4i+3j) into magnitude and that is as far as I got. Not a clue what to do next

 

[Hootenanny]

Homework Statement

Point A (2,3)m and point B (5,7)m are in a region where electric field is uniform and given by E=(4i+3j)N/C. What is potential diff Va-Vb?

Homework Equations

Not a clue

The Attempt at a Solution

I tried converting the (4i+3j) into magnitude and that is as far as I got. Not a clue what to do next

How is the electric field related to the potential?

 

[Bearbull24.5]

Isnt the potential the energy required to move a charge in an electric field?

 

[Hootenanny]

Isnt the potential the energy required to move a charge in an electric field?

It is indeed. So how does that relate to your problem?

 

[rwisz]

.

I would approach this problem by using the formula for potential difference, which is V = Ed, where V is the potential difference, E is the electric field, and d is the distance between the two points. In this case, the distance between point A and point B is given as 3 meters in the x-direction and 4 meters in the y-direction. Therefore, the total distance between the two points is d = √(3^2+4^2) = √25 = 5 meters.

Next, I would plug in the given values for the electric field (E = 4i+3j N/C) and the distance (d = 5m) into the formula for potential difference. This gives us:

V = (4i+3j) x 5 = 20i + 15j V

Therefore, the potential difference between point A and point B is 20 volts in the x-direction and 15 volts in the y-direction. It is important to note that the potential difference is a vector quantity, meaning it has both magnitude and direction. In this case, the direction of the potential difference is in the same direction as the electric field, since the two points are in the same direction as the electric field.

I hope this helps you understand how to approach this problem as a scientist. Remember to always use relevant equations and plug in the given values to find a solution.