An 800 V/m electric field is directed along the +x-axis. If the potential at x = 0 m is 2000 V, what is the potential at x = 2 m?

The potential difference at point x = 2m on the x-axis is 400 V.

The amount of energy transferred between two points in a circuit is measured by the potential difference between two points.

Given that the electric field along the x-axis is 800 V/m. Potential difference at x= 0 is 2000 V.

The electric field at two points along the x-axis can be given as,

Where E is the electric field, v1 is the potential difference at point x1 on the x-axis, and v2 is the potential difference at point x2 on the x-axis.

The potential difference v2 at point x2 = 2 m can be calculated by substituting the values in the above equation.

Hence we can conclude that the potential difference at point x = 2m on the x-axis is 400 V.

To know more about the potential difference, follow the link given below.

link.

400 V

Explanation:

Electric filed , E=800 V/m

We know that

Using the formula

Hence, the potential at x=2 m=400 V

the answer is a. hope this .

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s