Three point charges q,-2 q and 2 q are placed on x-axis at a distance x=0, x=3/4 R and x=R respectively from origin as shown. If q=2 × 10⁻⁶ C and R=2 cm, the magnitude of net force experienced by the charge -2 q is ___________ N.

Solution

Physics sir · 4 steps

Both outer charges pull -2q, but in opposite directions, so subtract the two forces, do not add them.

1. Directions first. q and 2q are both positive, so each one attracts -2q. The charge q at the origin pulls it to the left. The charge 2q at x = R pulls it to the right.
2. Distances. -2q is 3/4R = 1.5 cm from q and 1/4R = 0.5 cm from 2q. Use R = 0.02 m, so these are 0.015 m and 0.005 m.

   Fₙₑₜ = k(2q)(2q)/(R/4)² - k(q)(2q)/(3R/4)² = kq²/R²(64 - 32/9) = 544/9 kq²/R²

3. Put in numbers. kq²/R² = (9 × 10⁹ × (2 × 10⁻⁶)²)/(0.02)² = 90 N. So the pull to the right is 64 × 90 = 5760 N and the pull to the left is 32/9 × 90 = 320 N.
4. So. Fₙₑₜ = 5760 - 320 = 5440 N, directed towards 2q. The magnitude is 5440 N.

The common slip: 6080. Adding the two forces instead of subtracting them gives 5760 + 320 = 6080 N. The two attractions point in opposite directions.

The idea to keep. For a charge sitting between two others on a line, fix each force's direction from the signs before you calculate. Opposite pulls subtract, and 1/r² makes the nearer charge dominate.

Answer from the NTA final answer key. Solution by PrepWiser.