A uniform electric field is directed parallel to the +y axis. If a positive test charge begins at the origin and moves upward along the y axis, how does the electric potential vary, if at all?
1. The electric potential will decrease with increasing y.
2. The electric potential will increase with increasing y.
3. The electric potential will remain constant with increasing y.
4. Too little information is given to answer this question.

option 1 : the electric potential will decrease with an increase in y

Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below

V = kq/y

Where k = 1/4πε0

Where V = electric potential,

k = electric constant = 9×10^9,

y = distance of potential relative to a reference point, ε0 = permittivity of free space

q = magnitude of electronic charge = 1.609×10^-19 c

From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.

We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

yup its correct its: c) b b