Question

Two point charges $+3 \; \mu\text{C}$ and $+8 \; \mu\text{C}$ repel each other with a force of $40 \text{ N}$. If a charge of $-5 \; \mu\text{C}$ is added to each of them, then the force between them will become

• A. $-10 \text{ N}$
• B. $10 \text{ N}$
• C. $20 \text{ N}$
• D. $-20 \text{ N}$

Hint:

Keep an eye on the sign of your product to determine the direction of the force immediately! A positive product ($+ \times +$ or $- \times -$) means a repulsive force, while a negative product indicates attraction. Here, the final product is negative, meaning the force must be attractive (indicated by a negative sign in standard notation options).

Answer 1

The Correct Option is A

Step 1: Lay out the situation.
Charges $+3\ \mu\text{C}$ and $+8\ \mu\text{C}$ repel with $40\ \text{N}$. We add $-5\ \mu\text{C}$ to each and find the new force, sign included.
Step 2: Coulomb's law as a proportion.
$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$. With the separation $r$ fixed, $F \propto q_1 q_2$.
Step 3: Find the new charges.
$q_1' = 3 + (-5) = -2\ \mu\text{C}$ and $q_2' = 8 + (-5) = +3\ \mu\text{C}$.
Step 4: Set up the force ratio.
$\dfrac{F'}{F} = \dfrac{q_1' q_2'}{q_1 q_2} = \dfrac{(-2)(+3)}{(+3)(+8)}$.
Step 5: Evaluate the ratio.
$\dfrac{F'}{40} = \dfrac{-6}{24} = -\dfrac{1}{4}$.
Step 6: Solve for the new force.
$F' = 40\times\left(-\dfrac{1}{4}\right) = -10\ \text{N}$. The minus sign shows the force is now attractive, because one charge turned negative.
\[ \boxed{F' = -10\ \text{N (option 1)}} \]