Question

One point charge (q) each, is placed along a line at 3 different points \(x = 0\), \(x = 2 \, \text{nm}\) and \(x = 6 \, \text{nm}\). The force between two charges separated by 2 nm is 2 piconewton (pN). The magnitude of force (in pN) on the charge in the middle due to the other two charges is ..............

Hint:

When calculating the total force on a charge, add the forces from each individual interaction, considering directionality if needed.

Answer 1

Correct Answer: 1.5

We need to calculate the force on the middle charge at \(x = 2 \, \text{nm}\) due to the other two charges at \(x = 0\) and \(x = 6 \, \text{nm}\). According to Coulomb's Law, the force between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by:

\(F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\)

Here, the force between charges separated by 2 nm is 2 pN. We need to find:

• The force \(F_{01}\) between charges at \(x = 0\) and \(x = 2 \, \text{nm}\).
• The force \(F_{21}\) between charges at \(x = 2 \, \text{nm}\) and \(x = 6 \, \text{nm}\).

- The distance between \(x = 0\) and \(x = 2\) is 2 nm.

- The distance between \(x = 2\) and \(x = 6\) is 4 nm.

1. Since the force \(F = 2 \, \text{pN}\) for 2 nm separation, for \(F_{01}\) (same distance), the force is 2 pN.

2. For \(F_{21}\), at 4 nm, using inverse square law:

\(F_{21} = 2 \, \text{pN} \times \left(\frac{2}{4}\right)^2 = 2 \, \text{pN} \times \frac{1}{4} = 0.5 \, \text{pN}\)

The force directions for \(F_{01}\) and \(F_{21}\) are opposite (assuming positive direction is towards larger x). Therefore, net force \(F_{\text{net}}\) on middle charge:

\(F_{\text{net}} = F_{01} - F_{21} = 2 \, \text{pN} - 0.5 \, \text{pN} = 1.5 \, \text{pN}\)

This calculated net force = \(1.5 \, \text{pN}\)