Electric Field

Electric field tells you the force a small positive test charge would feel at a point in space. If you know the field, you know both the strength of the push or pull and its direction.

The definition is

$$E = \frac{F}{q}$$

where $F$ is the electric force on the test charge and $q$ is the test charge itself. This matters because the field belongs to the space around the source charges. The test charge only reveals what the field is doing there.

For a point charge $Q$ in vacuum, the field magnitude at distance $r$ is

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

where $k \approx 8.99 \times 10^9\ \mathrm{N \cdot m^2/C^2}$. The direction is away from a positive source charge and toward a negative source charge.

Electric Field Meaning In Plain Language

Electric field lets you describe electric influence without redoing the full force story for every new charge. Once you know the field at a point, you can predict the force on any charge placed there.

If the charge you place at that point is $q$, then

$$\vec{F} = q\vec{E}$$

So a larger charge feels a larger force in the same field. A negative charge feels a force opposite the field direction.

When The Electric Field Formula $E = k|Q|/r^2$ Works

The formula

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

is exact for a point charge in vacuum. It also works for points outside a spherically symmetric charge distribution, where the distribution acts as if all the charge were concentrated at the center.

If the charge distribution is extended and not spherically symmetric, one direct plug-in formula is usually not enough. In that case, the field is found by adding contributions from many small pieces of charge.

Worked Example: Find The Field And Then The Force

Suppose a point charge $Q = +2.0 \times 10^{-6}\ \mathrm{C}$ creates a field. Find the electric field at a point $0.50\ \mathrm{m}$ away, then find the force on a test charge $q = +3.0 \times 10^{-9}\ \mathrm{C}$ placed there.

Step 1: find the field magnitude.

$$E = k \frac{|Q|}{r^2}$$ $$E = (8.99 \times 10^9)\frac{2.0 \times 10^{-6}}{(0.50)^2}$$ $$E = (8.99 \times 10^9)\frac{2.0 \times 10^{-6}}{0.25} \approx 7.19 \times 10^4\ \mathrm{N/C}$$

Because the source charge is positive, the field points away from the source.

Step 2: use the field to find the force on the test charge.

$$F = (3.0 \times 10^{-9})(7.19 \times 10^4) \approx 2.16 \times 10^{-4}\ \mathrm{N}$$

The test charge is also positive, so the force points in the same direction as the field: away from the source. If the test charge were negative, the field would stay the same but the force would reverse direction.

Common Electric Field Mistakes

• Mixing up the source charge $Q$ with the test charge $q$.
• Forgetting that $E = F/q$ defines the field, while $F = qE$ gives the force on a particular charge.
• Using the inverse-square formula for situations that are not well modeled as a point charge or a spherically symmetric distribution.
• Dropping the direction. Electric field is a vector, not just a number.
• Forgetting that a negative test charge feels a force opposite the field direction.

Where Electric Field Is Used

Electric field is one of the core ideas in electrostatics. It appears in problems about point charges, charged spheres, capacitors, and the motion of charged particles.

It is also the bridge between Coulomb's law and later ideas such as electric potential and Gauss's law. If electric field feels clear, much of introductory electromagnetism becomes easier to organize.

Try A Similar Problem

Change only one thing in the example: make the test charge negative instead of positive. The field at that location stays the same, but the force direction changes. Try your own version with different charges and distances, and check whether you can predict the direction before you calculate the magnitude.