Increasing the distance from a point charge by 34% changes the magnitude of the electric field by what amount?
A) it decreases by 16%
B) it decreases by 34%
C) it decreases by 44%
D) it decreases by 80%
E) it stays the same

Question

Quizplus · Accepted Answer

The electric field (E) due to a point charge is given by Coulomb's law, $E = k \frac{Q}{r^2}$, where $k$ is Coulomb's constant, $Q$ is the charge, and $r$ is the distance from the charge. If the distance $r$ is increased by 34%, the new distance $r'$ is $1.34r$. Substituting $r'$ into Coulomb's law gives $E' = k \frac{Q}{{(1.34r)}^2}$. Simplifying, we find $E' = \frac{E}{{1.34}^2}$. Calculating $\frac{1}{{1.34}^2}$ gives approximately 0.56, meaning the electric field strength decreases to 56% of its original value, which is a 44% decrease.

Increasing the Distance from a Point Charge by 34% Changes