Electric Field — Point Charge (Worked Example)

Medium

Problem

A point charge of +2.0 μC sits alone in space. What is the magnitude of the electric field 0.30 m away from it? (k = 8.99 × 10⁹ N·m²/C²)

Given

Q = 2.0 μC = 2.0 × 10⁻⁶ C
r = 0.30 m
k = 8.99 × 10⁹ N·m²/C²

Find

The field magnitude E.

Principle

A charge Q creates a field E = kQ/r² at distance r — the force a 1 C test charge would feel there.

Formula

E = k\frac{Q}{r^2}

E = kQ/r²

Solution

Write the formula: $E = kQ/r^2$
Substitute: $E = (8.99 \times 10^9)(2.0 \times 10^{-6})/0.30^2$
Numerator: $1.798 \times 10^4$ ; denominator: $0.09$
Divide: $E = 2.00 \times 10^5\ \mathrm{N/C}$

2.00 × 10⁵ N/C, directed away from the charge

Unit check

N·m²/C² × C ÷ m² = N/C — correct.

What this means physically

2.00 × 10⁵ N/C — a strong field; a dust grain with 1 nC of charge would feel 0.2 N here. Fields fall as 1/r², so at 0.6 m it would be four times weaker.

Common mistake

Forgetting to convert μC to coulombs. A factor of 10⁶ error is the classic electric-field mistake — always convert micro to 10⁻⁶ first.