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2.3 - Electric Fields

What an Electric Field Is

The electric field (E⃗\vec{E}) is the electric force per unit charge at a point in space.
E⃗=F⃗Eq\vec{E} = \frac{\vec{F}_E}{q}
Rearranged, that gives:
F⃗E=qE⃗\vec{F}_E = q\vec{E}
where: - E⃗\vec{E} = electric field (N/C or V/m) - F⃗E\vec{F}_E = electric force on a test charge (N) - qq = test charge (C)
The units are N/C (newtons per coulomb) or V/m (volts per meter), and the two are equivalent.
The main idea is that the electric field exists at every point in space around a charge, whether or not there’s another charge sitting there to “feel” it.

Test Charge

A test charge is a small positive charge you use to probe the electric field without disturbing it much. It needs to be: - Small in magnitude, so it doesn’t affect the field. - Treated as positive by convention.

Electric Field from a Point Charge

For a point charge QQ, the field at a distance rr is:
E=k∣Q∣r2E = k\frac{|Q|}{r^2}
The direction depends on the sign: - For a positive charge (+Q+Q), the field points away, radially outward. - For a negative charge (−Q-Q), the field points toward it, radially inward.
A quick way to remember it: positive pushes away, negative pulls in.

Force on a Charge in an Electric Field

For a charge qq sitting in a field E⃗\vec{E}:
F⃗E=qE⃗\vec{F}_E = q\vec{E}
Again, the direction depends on the sign: - A positive charge (q>0q > 0) feels a force in the same direction as E⃗\vec{E}. - A negative charge (q<0q < 0) feels a force in the opposite direction to E⃗\vec{E}.

Superposition of Electric Fields

The net electric field is the vector sum of all the individual fields:
E⃗net=E⃗1+E⃗2+E⃗3+...\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + ...
The process is: 1. Find the field from each charge on its own. 2. Work out the direction of each field. 3. Add them as vectors, using either components or geometry.
The key point is that fields from different sources add independently.
On the AP exam you’ll only have to calculate the net field from four or fewer charges, unless the setup is very symmetric.

Ways to Represent Electric Fields

1. Vector Field Maps

An electric field map shows field vectors at many points in space: - The arrow’s direction is the field direction. - The arrow’s length is the field strength, so longer means stronger.

2. Electric Field Lines

Field lines are a simplified model that shows field direction and relative strength. The rules for drawing them are: 1. Lines start on positive charges, or come in from infinity. 2. Lines end on negative charges, or go out to infinity. 3. Lines never cross. 4. Lines are tangent to the field direction at every point. 5. The density of lines is proportional to field strength, so closer lines mean a stronger field. 6. Lines meet conductor surfaces perpendicularly.
2.3 - Electric Fields
2.3 - Electric Fields

Common Configurations

Parallel plates (uniform field): - The field lines are parallel and evenly spaced between the plates. - The field is uniform, so E=constantE = \text{constant} between the plates. - It points from the positive plate to the negative plate.
Two like charges (both + or both −): - The lines push away from each other. - There’s a weak field region between the charges.
Two opposite charges (a dipole): - Lines run from + to −. - The field is strong between the charges.

Conductors in Electrostatic Equilibrium

Electrostatic equilibrium means there’s no net motion of charges and the system is stable.
Conductors in equilibrium have four properties worth memorizing:
  1. The field is zero inside the conductor.
      • The free charges rearrange until there’s no field left.
      • Any field inside would push charges around, which wouldn’t be equilibrium.
  1. Excess charge sits on the surface.
      • The charges repel each other out to the outer surface.
      • The interior stays neutral.
  1. The field is perpendicular to the surface at the boundary.
      • Any parallel component would drive surface charges to move.
      • So it has to be perpendicular for equilibrium.
  1. Charge piles up where the surface is most curved.
      • Sharp points get high surface charge density, which makes a strong field.
      • Flat regions get low surface charge density and a weak field.
      • This is exactly why lightning rods have sharp tips.

Spherical Conductor

Outside a spherical conductor (radius RR, charge QQ), for r>Rr > R:
E=kQr2E = k\frac{Q}{r^2}
This is the same as a point charge with charge QQ sitting at the center.
Inside a spherical conductor (r<Rr < R):
E=0E = 0

Insulators (Dielectrics)

In insulators, charges can’t move around freely.
In electrostatic equilibrium: - Excess charge spreads throughout the whole volume, not just the surface. - The electric field can be nonzero inside. - Polarization can happen, meaning charge separation at the molecular level.
For AP Physics 2, we only do qualitative analysis of fields inside insulators.

Practice Problem 1: Field from a Point Charge

Problem: A charge Q=+5.0×10−6Q = +5.0 \times 10^{-6} C creates an electric field. Find the field’s magnitude and direction at a point 0.30 m away.
Solution:
E=k∣Q∣r2=(8.99×109)5.0×10−6(0.30)2E = k\frac{|Q|}{r^2} = (8.99 \times 10^9)\frac{5.0 \times 10^{-6}}{(0.30)^2}E=(8.99×109)5.0×10−60.09=5.0×105 N/CE = (8.99 \times 10^9)\frac{5.0 \times 10^{-6}}{0.09} = 5.0 \times 10^5 \text{ N/C}
Direction: away from the positive charge, radially outward.
Answer: 5.0×1055.0 \times 10^5 N/C, radially outward.

Practice Problem 2: Superposition

Problem: Two charges: q1=+4.0q_1 = +4.0 μC at the origin, q2=−2.0q_2 = -2.0 μC at x=0.60x = 0.60 m. Find the net electric field at x=0.30x = 0.30 m, the midpoint.
Solution:
Field from q1q_1 (0.30 m away, on the right):
E1=k∣q1∣r12=(8.99×109)4.0×10−6(0.30)2=4.0×105 N/CE_1 = k\frac{|q_1|}{r_1^2} = (8.99 \times 10^9)\frac{4.0 \times 10^{-6}}{(0.30)^2} = 4.0 \times 10^5 \text{ N/C}
Direction: away from +q1+q_1, so it points right (+x).
Field from q2q_2 (0.30 m away, on the left):
E2=k∣q2∣r22=(8.99×109)2.0×10−6(0.30)2=2.0×105 N/CE_2 = k\frac{|q_2|}{r_2^2} = (8.99 \times 10^9)\frac{2.0 \times 10^{-6}}{(0.30)^2} = 2.0 \times 10^5 \text{ N/C}
Direction: toward −q2-q_2, so it also points right (+x).
Net field, since both point right:
Enet=E1+E2=4.0×105+2.0×105=6.0×105 N/CE_{net} = E_1 + E_2 = 4.0 \times 10^5 + 2.0 \times 10^5 = 6.0 \times 10^5 \text{ N/C}
Answer: 6.0×1056.0 \times 10^5 N/C to the right.

Practice Problem 3: Force in a Field

Problem: An electron is placed in a uniform electric field of magnitude 2.0×1042.0 \times 10^4 N/C pointing downward. Find the force on the electron, both magnitude and direction.
Solution:
FE=∣q∣E=(1.60×10−19)(2.0×104)=3.2×10−15 NF_E = |q|E = (1.60 \times 10^{-19})(2.0 \times 10^4) = 3.2 \times 10^{-15} \text{ N}
Direction: the electron is negative, so the force is opposite to the field. The field points down, so the force points up.
Answer: 3.2×10−153.2 \times 10^{-15} N upward.

Practice Problem 4: Field Off to the Side of a Dipole

Problem: A +2.0+2.0 nC charge and a −2.0-2.0 nC charge are 8.0 cm apart. Point PP is 3.0 cm above the midpoint between them, so it’s 5.0 cm from each charge. Find the net electric field at PP.
2.3 - Electric Fields
Solution:
Both charges have the same magnitude and are the same distance from PP (r=0.050r = 0.050 m), so their fields have the same magnitude:
E1=E2=k∣q∣r2=(8.99×109)2.0×10−9(0.050)2=7.2×103 N/CE_1 = E_2 = k\frac{|q|}{r^2} = (8.99 \times 10^9)\frac{2.0 \times 10^{-9}}{(0.050)^2} = 7.2 \times 10^3 \text{ N/C}
Directions: - E⃗1\vec{E}_1 points away from the positive charge, so up and to the right, along the line from +q+q through PP. - E⃗2\vec{E}_2 points toward the negative charge, so down and to the right, along the line from PP to −q-q.
Break each one into components. Each line makes an angle θ\theta with the horizontal, and the 3-4-5 triangle gives cos⁡θ=4.0/5.0=0.80\cos\theta = 4.0/5.0 = 0.80 and sin⁡θ=3.0/5.0=0.60\sin\theta = 3.0/5.0 = 0.60:
Ex=E1cos⁡θ+E2cos⁡θ=2(7.2×103)(0.80)=1.2×104 N/CE_x = E_1\cos\theta + E_2\cos\theta = 2(7.2 \times 10^3)(0.80) = 1.2 \times 10^4 \text{ N/C}Ey=E1sin⁡θ−E2sin⁡θ=0E_y = E_1\sin\theta - E_2\sin\theta = 0
Answer: 1.2×1041.2 \times 10^4 N/C, pointing right (parallel to the line from the positive charge to the negative charge).
The y-components cancel by symmetry, but the x-components add. That’s why the field everywhere on this midline points the same way, from the + side toward the − side.

Practice Problem 5: Where Is the Field Zero?

Problem: A +1.0+1.0 nC charge sits at x=0x = 0 and a +4.0+4.0 nC charge sits at x=0.90x = 0.90 m. Where on the x-axis is the net electric field zero?
Solution:
First figure out which region it can be in: - To the left of both charges, or to the right of both, the two fields point the same way, so they can’t cancel. - Between the charges, the fields point in opposite directions (each one points away from its own positive charge), so they can cancel there. - The zero point has to be closer to the smaller charge, since being closer is the only way its field can match the bigger charge’s field.
Call the zero point xx and set the two magnitudes equal:
k1.0×10−9x2=k4.0×10−9(0.90−x)2k\frac{1.0 \times 10^{-9}}{x^2} = k\frac{4.0 \times 10^{-9}}{(0.90 - x)^2}
Cancel kk and 10−910^{-9}, then take the square root of both sides (both distances are positive):
0.90−xx=4.0=2.0⇒0.90−x=2.0x⇒x=0.30 m\frac{0.90 - x}{x} = \sqrt{4.0} = 2.0 \quad\Rightarrow\quad 0.90 - x = 2.0x \quad\Rightarrow\quad x = 0.30 \text{ m}
Check: E1=(8.99×109)(1.0×10−9)/(0.30)2=1.0×102E_1 = (8.99 \times 10^9)(1.0 \times 10^{-9})/(0.30)^2 = 1.0 \times 10^2 N/C to the right, and E2=(8.99×109)(4.0×10−9)/(0.60)2=1.0×102E_2 = (8.99 \times 10^9)(4.0 \times 10^{-9})/(0.60)^2 = 1.0 \times 10^2 N/C to the left. ✓
Answer: x=0.30x = 0.30 m, between the charges and closer to the smaller one.
If the two charges had opposite signs, the zero point would be outside them instead, on the side of the smaller charge.

Common Mistakes

  • Confusing field and force. The field exists everywhere, but you need a charge present to get a force.
  • Getting the direction wrong for a negative charge. A negative charge feels a force opposite to the field direction.
  • Adding magnitudes instead of vectors. You have to account for direction when finding the net field.
  • Forgetting the r2r^2 in the denominator. Distance is squared in Coulomb-type equations.
  • Thinking field lines are paths. Field lines show the direction of the force on a positive charge at each point, not the path a charge follows. A charge only moves along a field line in special cases, like a charge that starts from rest in a uniform field.
  • Assuming the field inside a conductor is nonzero. In equilibrium, E=0E = 0 inside a conductor, always.
  • Getting the perpendicular wrong. At a conductor’s surface, the field is perpendicular (90°), not tangent.
  • Thinking excess charge spreads through the whole conductor. For a conductor in equilibrium, excess charge is only on the surface.

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