One point charge (q) each, is placed along a line at 3 different points x = 0, x = 2 nm and x = 6 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 ________.
Problem Setup
Three identical point charges q sit along a line at x = 0, x = 2 nm (middle), and x = 6 nm. The force between any two charges separated by 2 nm equals 2 pN, following Coulomb’s law: F = k \frac{q^2}{r^2}, where k ≈ 9 × 10^9 Nm²/C² and r = 2 × 10^{-9} m.
🔍 Charge Configuration
Charge 1: x = 0 nm → Charge 2 (middle): x = 2 nm → Charge 3: x = 6 nm
Distances: 2 nm (left) | 4 nm (right)
Since all charges have the same sign (implied by “one point charge q each” and magnitude focus), forces are repulsive.
Force Calculations
The middle charge at x = 2 nm experiences:
- Force from left charge (x = 0): distance = 2 nm, so F_L = 2 pN (to the right).
- Force from right charge (x = 6 nm): distance = 4 nm = 2r, so F_R = k \frac{q^2}{(4 × 10^{-9})^2} = \frac{k q^2}{(2r)^2} = \frac{F}{4} = \frac{2}{4} = 0.5 pN (to the left).
Net Force by Superposition
Net force uses the superposition principle: vector sum of individual forces.
SEO-Optimized Article Content
The force on middle charge in a system of three identical point charges placed at x = 0, x = 2 nm, and x = 6 nm presents a classic application of Coulomb’s law and the superposition principle, ideal for competitive exams like CSIR NET physics and JEE.
Step-by-Step Solution
- Left Force (F_L): Separation = 2 nm → F_L = 2 pN (repulsive, rightward).
- Right Force (F_R): Separation = 4 nm → F_R = 2 × \left(\frac{2}{4}\right)^2 = 2 × 0.25 = 0.5 pN (repulsive, leftward).
- Net Force: |F_net| = 2 – 0.5 = 1.5 pN (rightward).
Common Exam Traps
- ❌ Wrong: Adding magnitudes = 2.5 pN (ignores direction)
- ❌ Wrong: Using linear distance scaling
- ✅ Correct: 1.5 pN using 1/r^2 law
Practice similar problems for CSIR NET exam success! Master point charges force calculation using superposition principle.
