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: Understanding the Question:
Two point charges separated by a fixed distance experience a known repulsive electrostatic force. We modify the system by adding a specific negative charge to both particles. We need to determine the new electrostatic force between them, tracking both its magnitude and its direction (repulsive vs. attractive).

Step 2: Key Formula or Approach:
According to Coulomb's Law, the electrostatic force $F$ between two static point charges is directly proportional to the product of their charge magnitudes:
$$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \implies F \propto q_1 \cdot q_2$$ Since the spatial separation distance $r$ remains unchanged, we can establish a direct ratio comparison between the final force ($F'$) and initial force ($F$):
$$\frac{F'}{F} = \frac{q_1' \cdot q_2'}{q_1 \cdot q_2}$$

Step 3: Detailed Explanation:
Let's analyze the initial and final states of the charges:

Initial Charges: $q_1 = +3 \; \mu\text{C}$, $q_2 = +8 \; \mu\text{C}$

Initial Force: $F = 40 \text{ N}$ (positive sign represents mutual repulsion)
Now, add a charge of $-5 \; \mu\text{C}$ to each point source to determine the modified values:
$$q_1' = +3 \; \mu\text{C} + (-5 \; \mu\text{C}) = -2 \; \mu\text{C}$$ $$q_2' = +8 \; \mu\text{C} + (-5 \; \mu\text{C}) = +3 \; \mu\text{C}$$ Set up the force ratio equation using our algebraic proportionality relation:
$$\frac{F'}{40} = \frac{(-2) \times (+3)}{(+3) \times (+8)}$$ $$\frac{F'}{40} = \frac{-6}{24} = -\frac{1}{4}$$ Isolate the new force $F'$ by multiplying both sides by 40:
$$F' = 40 \times \left(-\frac{1}{4}\right) = -10 \text{ N}$$ The negative sign indicates that the nature of the force has converted from repulsive to attractive, because the charges now have opposite signs.

Step 4: Final Answer:
The force between the charges becomes $-10 \text{ N}$, which corresponds to option (A).