The electric potential at a certain distance from a point charge can be represented by v. what is the value of the electric potential at twice the distance from the point charge?
the electric potential at a certain distance from a point charge can be represented by v. what is the value of the electric potential at twice the distance from the point charge?
at twice the distance, the electric potential is 4v.
at twice the distance, the electric potential is v/2.
at twice the distance, the electric potential is v/4.
at twice the distance, the electric potential is 2v.
at twice the distance, the electric potential remains v.

At twice the distance, the electric potential is V/2.

Explanation:

The electric potential at a certain distance d from a point charge q can be represented by V:

V = kq/d 1

Where q = charge ,d = distance between them, k = coulomb's constant.

When the distance is doubled d1 = 2d

Let V1 represent the electric potential after the distance have been doubled.

V1 = kq/2d = (kq/d)/2

V1 = V/2

Therefore, at twice the distance the electric potential is halved V1 = V/2

c. magnitude and direction

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