Question

The electric potential at a distance $r$ from a point charge $q$ is proportional to

• A. $1/r$
• B. $1/r^2$
• C. $r$
• D. $r^2$

Hint:

Remember that the electric potential due to a point charge decreases as you move away from it in an inverse relationship with the distance.

Answer 1

The Correct Option is A

Step 1: Recall the potential formula.
The electric potential at distance $r$ from a point charge $q$ is \[ V = k\frac{q}{r} \] where $k$ is Coulomb's constant.

Step 2: Spot the dependence.
In this formula $r$ sits in the bottom of the fraction to the first power. So \[ V \propto \frac{1}{r} \]

Step 3: Test by doubling distance.
If we double $r$, the potential becomes \[ V' = k\frac{q}{2r} = \frac{1}{2}V \] So twice the distance gives half the potential. This matches the inverse relation.

Step 4: Compare with field.
Note the electric field falls as $1/r^2$, but the potential falls more slowly as $1/r$. Do not mix the two.

Step 5: Rule out the others.
Choices like $1/r^2$, $r$, or $r^2$ do not match the simple $1/r$ behavior of the potential.

Step 6: State the answer.
The potential is proportional to the inverse of the distance. \[ \boxed{V \propto \frac{1}{r}} \]