recall and use F = qE for the force on a charge in an electric field

Electric Fields and Field Lines

Learning Objective

Recall and use the relationship F = qE to determine the force acting on a charge placed in an electric field.

Key Concepts

• Definition of electric field
• Direction of field lines
• Magnitude of the field
• Superposition of fields
• Using \$F = qE\$ in calculations

Definition of Electric Field

The electric field \$\mathbf{E}\$ at a point in space is defined as the force \$\mathbf{F}\$ experienced by a positive test charge \$q_{0}\$ placed at that point, divided by the magnitude of the test charge:

\$\mathbf{E} = \frac{\mathbf{F}}{q_{0}}\$

Units: newtons per coulomb (N C⁻¹) or volts per metre (V m⁻¹).

Field Lines

Field lines are a visual tool to represent the direction and relative strength of an electric field.

• Lines originate on positive charges and terminate on negative charges.
• The tangent to a field line at any point gives the direction of \$\mathbf{E}\$ there.
• Density of lines (lines per unit area) is proportional to the magnitude of the field.

Using \$F = qE\$

Once the electric field at a location is known, the force on any charge \$q\$ placed there is given by:

\$\mathbf{F} = q\,\mathbf{E}\$

Important points:

1. If \$q\$ is positive, the force is in the same direction as the field.
2. If \$q\$ is negative, the force is opposite to the field direction.
3. The magnitude is \$F = |q|E\$.

Example: Force on a Charge in a Uniform Field

Consider a uniform electric field of magnitude \$E = 5.0 \times 10^{3}\,\text{N C}^{-1}\$ directed horizontally to the right. Find the force on a charge \$q = -2.0\,\mu\text{C}\$.

1. Convert the charge: \$q = -2.0 \times 10^{-6}\,\text{C}\$.
2. Apply \$F = qE\$:

   \$F = (-2.0 \times 10^{-6}\,\text{C})(5.0 \times 10^{3}\,\text{N C}^{-1}) = -1.0 \times 10^{-2}\,\text{N}\$

3. The negative sign indicates the force is opposite to the field direction, i.e., to the left.

Superposition of Electric Fields

When multiple charges are present, the net electric field at a point is the vector sum of the fields due to each charge:

\$\mathbf{E}{\text{net}} = \sum{i}\mathbf{E}_{i}\$

Consequently, the force on a test charge is:

\$\mathbf{F} = q\,\mathbf{E}_{\text{net}}\$

Table: Field Direction for Common Charge Configurations

Practice Questions

1. Two point charges, \$+3\;\mu\text{C}\$ and \$-3\;\mu\text{C}\$, are 0.10 m apart. Calculate the magnitude and direction of the electric field at the midpoint.
2. A uniform field of \$2.0\times10^{4}\,\text{N C}^{-1}\$ points upward. What force does it exert on a proton (\$q = +1.60\times10^{-19}\,\text{C}\$)?
3. Three charges are placed at the vertices of an equilateral triangle of side \$0.05\,\$m: \$+2\;\mu\text{C}\$, \$+2\;\mu\text{C}\$, and \$-4\;\mu\text{C}\$. Determine the net electric field at the centre of the triangle.

Summary

• The electric field \$\mathbf{E}\$ describes the force per unit positive charge.
• Field lines give a convenient visual representation: direction = tangent, strength = density.
• Use \$ \mathbf{F}=q\mathbf{E}\$, remembering the sign of \$q\$ determines the direction of the force.
• For multiple sources, add fields vectorially before applying \$F = qE\$.