Question

question 15 of 25
how does the electric force between two charged particles change if the distance between them is reduced by a factor of 2?
a. it is increased by a factor of 2.
b. it is reduced by a factor of 2.
c. it is reduced by a factor of 4.
d. it is increased by a factor of 4.

Explanation:

Step1: Recall Coulomb's law

The electric force between two charged particles is given by $F = k\frac{q_1q_2}{r^2}$, where $k$ is the Coulomb's constant, $q_1$ and $q_2$ are the charges of the two particles, and $r$ is the distance between them.

Step2: Consider the new - distance

Let the original distance be $r$ and the new distance $r'=\frac{r}{2}$. The new force $F'$ is $F'=k\frac{q_1q_2}{r'^2}=k\frac{q_1q_2}{(\frac{r}{2})^2}$.

Step3: Simplify the expression for the new force

$F' = k\frac{q_1q_2}{\frac{r^2}{4}}=4\times k\frac{q_1q_2}{r^2}$. Since $F = k\frac{q_1q_2}{r^2}$, we have $F' = 4F$.

Answer:

D. It is increased by a factor of 4.