Answer:
Explanation:
Consider the axis diagram attached.
Given:
Ey = Ez = 0
Eₓ = - 4x N/C · m
Since electric field is in x direction, potential difference would be:
Here we integrate between limits 0 and 4.40 which is distance between A and B along x-axis.
![V_{b} - V_{a} = -4 \left[\begin{array}{ccc}\frac{x^{2} }{2} \end{array}\right]^{4.40}_{0}](https://tex.z-dn.net/?f=V_%7Bb%7D%20-%20V_%7Ba%7D%20%3D%20-4%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D%5E%7B4.40%7D_%7B0%7D)
Answer:
Explanation:
First let's find the electric potential using y = 22.5:
Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:
E=Eq,
where F is the electrostatic force (or Coulomb force) exerted on a positive test charge q.
many increases and decreases but declined overall
Answer:
False.
Explanation:
Lets assume our positive direction to the right (this reasoning works for any direction). A negative velocity would then be then directed to the left. If it varies as such that it aproaches to zero, it means that the variation is directed to the right, and that is where the direction of the acceleration must be pointing. In other words, its losing its velocity, so the acceleration must point opposite to the velocity. Then it means the acceleration is positive.
