Electric fields

Definition and Direction of Electric Fields

Electric fields originate from charged objects and fill the surrounding space, exerting forces on any other charges placed in that region.

Key Equations

Electric field from force on a test charge:

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

Variables:

• $\vec{E}$ = electric field vector (N/C)
• ${\vec{F}}_{E}$ = electric force on the test charge (N)
• $q$ = charge of the test charge (C)

SI unit: N/C (equivalently V/m)

Rearrangement:

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

ProportionalityThe electric field is directly proportional to the force on the test charge. Doubling the force doubles the measured field.

Reference sheet status: On the reference sheet.

An electric field is a region where a charged object experiences an electric force. The field exists whether or not a second charge is present to "feel" it.

To measure the field at any point, place a small positive test charge $q$ there. The test charge must be small enough that it does not significantly affect the field it is measuring. The electric field at that point equals the electric force on the test charge divided by the charge of the test charge.

Direction follows a simple rule: the electric field points away from positive charges and toward negative charges. Because the field is defined using a positive test charge:

• the force on a positive test charge is in the same direction as the electric field.
• a negative charge placed in the same field experiences a force opposite to the field direction.

MisconceptionStudents often assume the electric field exists only when a test charge is present. The field exists at every point around a source charge regardless. The test charge is just a measurement tool.

Exam TipAlways define E⃗ as force per unit charge, not just force.

Superposition and Electric Field Maps

The net electric field at any point equals the vector sum of the individual electric fields from every source charge. This is the principle of superposition applied to electric fields.

When multiple charges are present, each charge creates its own field independently. To find the total field at a point, calculate each individual field vector, then add them using vector addition:

• In one dimension, assign a positive direction and add components with appropriate signs.
• In two dimensions, resolve each field vector into perpendicular components ($\hat{i}$ and $\hat{j}$), sum the components separately, then recombine.

Reading vector notation: The symbols $\hat{i}$, $\hat{j}$, and $\hat{k}$ are unit vectors pointing along the $x$-, $y$-, and $z$-axes. Writing $\vec{E}={E}_{x}\hat{i}+{E}_{y}\hat{j}$ means the field has component ${E}_{x}$ along the $x$-axis and ${E}_{y}$ along the $y$-axis.

Electric field maps use vectors drawn at many points to depict the magnitude and direction of the field. Longer arrows represent stronger fields. Electric field line diagrams are simplified models of these vector maps. Lines leave positive charges and enter negative charges. Where lines are close together, the field is strong. Where they spread apart, the field is weak.

Examiner InsightFRQs often ask students to sketch or interpret field line diagrams near two charges. Lines must never cross, because the field has only one direction at each point.

Exam TipState "the net field is the vector sum of individual fields" whenever superposition is involved.

Worked Example: Superposition of Two Point Charges

Scenario

Two charges, ${q}_{1}=+3.0\text{ μC}$ and ${q}_{2}=-3.0\text{ μC}$, are separated by 0.40 m along the $x$-axis. ${q}_{1}$ is at the origin and ${q}_{2}$ is at $x=0.40$ m. Determine the net electric field at the midpoint between them.

Step-by-step solution:

Choose the positive direction as the $+x$ direction (from ${q}_{1}$ toward ${q}_{2}$).

The midpoint is at $x=0.20$ m, a distance $r=0.20$ m from each charge.

Unit conversion:

$${q}_{1}=3.0\times {10}^{-6}\text{ C}$$

$${q}_{2}=3.0\times {10}^{-6}\text{ C (magnitude)}$$

Equation used

$$E=\frac{kq}{{r}^{2}}$$

Field from ${q}_{1}$ at the midpoint (positive charge, so field points away, in the $+x$ direction):

$${E}_{1}=\frac{(8.99\times {10}^{9})(3.0\times {10}^{-6})}{(0.20{)}^{2}}$$

$${E}_{1}=\frac{2.697\times {10}^{4}}{0.040}$$

$${E}_{1}=6.74\times {10}^{5}\text{ N/C in the }+x\text{ direction}$$

Field from ${q}_{2}$ at the midpoint (negative charge, so field points toward it, also in the $+x$ direction):

$${E}_{2}=\frac{(8.99\times {10}^{9})(3.0\times {10}^{-6})}{(0.20{)}^{2}}$$

$${E}_{2}=6.74\times {10}^{5}\text{ N/C in the }+x\text{ direction}$$

Net field by superposition:

$${E}_{\text{net}}={E}_{1}+{E}_{2}=6.74\times {10}^{5}+6.74\times {10}^{5}$$

$${E}_{\text{net}}=1.35\times {10}^{6}\text{ N/C in the }+x\text{ direction}$$

Interpretation

Both fields point in the same direction at the midpoint of a dipole. The fields reinforce along the axis connecting opposite charges.

Fields in Conductors at Electrostatic Equilibrium

Inside a conductor in electrostatic equilibrium [no net motion of charge], the electric field is zero everywhere within the conducting material.

This fact follows from the nature of conductors. Charge carriers in a conductor are free to move. If an electric field existed inside the conductor, charges would accelerate. That motion contradicts the assumption of electrostatic equilibrium. So the internal field must be zero.

All excess charge on a conductor resides on its outer surface. The charges repel each other and spread to the surface to maximize their separation. At the surface of a charged conductor, the electric field is perpendicular to the surface at every point. If the field had a component parallel to the surface, surface charges would move along the surface. That again contradicts equilibrium.

For a special and important case: the field outside an isolated sphere with a spherically symmetric charge distribution equals that of a point charge located at the center of the sphere. This means at any point outside the sphere, the field magnitude is:

$$E=\frac{kQ}{{r}^{2}}$$

where $Q$ is the total charge and $r$ is the distance from the center. This result holds for $r\ge R$, where $R$ is the radius of the sphere.

MisconceptionStudents sometimes think the field inside a hollow conductor is always zero only when the conductor is solid. The field is zero throughout the conducting material itself, and also zero in any hollow cavity (provided no charge is inside the cavity).

Exam TipState "the conductor is in electrostatic equilibrium" before claiming zero internal field.

Fields in Insulators at Electrostatic Equilibrium

In an insulator in electrostatic equilibrium, excess charge may be distributed throughout the interior as well as on the surface, and the electric field inside may be nonzero.

This behavior differs sharply from conductors. In an insulator, charge carriers are not free to move. Whatever charge distribution is placed inside the material stays fixed. Because internal charges can exist, they can produce a nonzero electric field within the insulator itself. The field inside depends on the specific charge distribution.

Examiner InsightFRQs often present a charged insulating sphere with uniform volume charge density and ask for the field at an interior point. The field inside increases linearly with distance from the center in that case. Contrast this with the zero field inside a conductor.

Exam TipWhen comparing conductors and insulators, always state why charges behave differently — mobility vs. immobility.