A uniform electric field is one where the electric field strength (E) is the same at every point in space. Think of it like a set of parallel invisible lines that push charged particles in the same direction, just like a steady wind blowing across a field of flags. The field lines are straight and evenly spaced, indicating that the force on a charge would be the same no matter where it is in the field.
Coulomb’s law tells us how strong the force is between two point charges. It’s the “rule of thumb” for electric forces in free space.
Formula:
\$F = \frac{Q1 Q2}{4 \pi \varepsilon_0 r^2}\$
Remember: The force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The sign of the force (attractive or repulsive) depends on whether the charges are like or unlike.
Solution:
First, convert microcoulombs to coulombs: \$3\,\mu\text{C} = 3\times10^{-6}\,\text{C}\$, \$-2\,\mu\text{C} = -2\times10^{-6}\,\text{C}\$.
Plug into Coulomb’s law:
\$F = \frac{(3\times10^{-6})(-2\times10^{-6})}{4\pi(8.85\times10^{-12})(0.05)^2}\$
\$F \approx -\frac{6\times10^{-12}}{4\pi(8.85\times10^{-12})(0.0025)} \approx -\frac{6\times10^{-12}}{2.78\times10^{-10}} \approx -0.0216\,\text{N}\$
The negative sign indicates an attractive force (since the charges are opposite). The magnitude is about \$0.022\,\text{N}\$.
| \$Q_1\$ (C) | \$Q_2\$ (C) | \$r\$ (m) | \$F\$ (N) |
|---|---|---|---|
| \$+1\times10^{-6}\$ | \$+1\times10^{-6}\$ | 0.01 | \$0.009\$ |
| \$-2\times10^{-6}\$ | \$+3\times10^{-6}\$ | 0.02 | \$-0.015\$ |
| \$+5\times10^{-6}\$ | \$-5\times10^{-6}\$ | 0.05 | \$-0.022\$ |