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Electric Fields

Rather than thinking of one charge exerting a direct "action at a distance" force on another, it is far more powerful to introduce the electric field: a property of space itself. A source charge creates a field everywhere around it; a second charge then responds to the field at its location. This field concept becomes indispensable when we get to time-varying fields and electromagnetic waves.

Key Concepts

Electric Field Definition
The electric field E⃗\vec{E} at a point is defined as the force per unit positive test charge placed at that point: E⃗=F⃗/q0\vec{E} = \vec{F}/q_0. Units: N/C (equivalently V/m). The test charge q0q_0 must be small enough not to disturb the source charges.
Field of a Point Charge
A point charge qq creates a field E=k∣q∣/r2E = k|q|/r^2 at distance rr. The field points radially outward from a positive charge and inward toward a negative charge.
Electric Field Lines
Field lines are a visual representation: they point in the direction of E⃗\vec{E}, originate on positive charges and terminate on negative charges, never cross, and are denser where the field is stronger.
Superposition of Fields
The total electric field at a point due to several charges is the vector sum of the individual fields: E⃗net=E⃗1+E⃗2+⋯\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 + \cdots. For continuous distributions, this sum becomes an integral.
Electric Dipole
Two equal and opposite charges +q+q and βˆ’q-q separated by distance dd. The dipole moment is pβƒ—=qdβƒ—\vec{p} = q\vec{d} (pointing from βˆ’q-q to +q+q), with magnitude p=qdp = qd. Dipoles in a uniform external field experience a torque Ο„=pEsin⁑θ\tau = pE\sin\theta that tends to align them with the field.

Key Equations

Electric field from force
E⃗=F⃗q0\vec{E} = \frac{\vec{F}}{q_0}
Definition of the electric field at a point in terms of force on a positive test charge qβ‚€.
Field of a point charge
E=k∣q∣r2E = k\frac{|q|}{r^2}
Magnitude of the field at distance r from point charge q. Direction: away from positive q, toward negative q.
Electric dipole moment
p=qdp = qd
Magnitude of the dipole moment. p points from βˆ’q to +q. Torque in field: Ο„ = pE sin ΞΈ.
Torque on dipole in uniform field
Ο„=pEsin⁑θ\tau = pE\sin\theta
Torque on a dipole of moment p in field E; ΞΈ is the angle between p and E. Tends to align the dipole with the field.
Worked Example

Net Electric Field Between Two Opposite Charges

Problem

Charge q1=+5q_1 = +5 nC is at the origin and q2=βˆ’5q_2 = -5 nC is at x=0.20x = 0.20 m. Find the electric field at the midpoint x=0.10x = 0.10 m.

Solution

Field from q1q_1 (positive, at origin) points away from it β€” in the +x+x direction at x=0.10x = 0.10 m:

E1=kq1r2=(8.99Γ—109)5Γ—10βˆ’9(0.10)2=4495Β N/CE_1 = k\frac{q_1}{r^2} = (8.99\times10^9)\frac{5\times10^{-9}}{(0.10)^2} = 4495 \text{ N/C}

Field from q2q_2 (negative, at x=0.20x = 0.20 m) points toward it β€” also in the +x+x direction at the midpoint:

E2=k∣q2∣r2=(8.99Γ—109)5Γ—10βˆ’9(0.10)2=4495Β N/CE_2 = k\frac{|q_2|}{r^2} = (8.99\times10^9)\frac{5\times10^{-9}}{(0.10)^2} = 4495 \text{ N/C}

Both fields point in the same direction (+x), so they add:

Enet=E1+E2=4495+4495=8990Β N/CE_{\text{net}} = E_1 + E_2 = 4495 + 4495 = 8990 \text{ N/C}
Answer E_net = 8990 N/C in the +x direction.
Practice

Exercises

7 problems
1 of 7

A point charge q=+8q = +8 nC. What is the magnitude of the electric field (in N/C) at a distance r=0.40r = 0.40 m from it?

N/C
2 of 7

A charge q0=+3 μCq_0 = +3\,\mu\text{C} is placed in a uniform electric field E=2.0Γ—105E = 2.0\times10^5 N/C. What is the magnitude of the force (in N) on it?

N
3 of 7

Charges q1=+5q_1 = +5 nC (at x=0x = 0) and q2=βˆ’5q_2 = -5 nC (at x=0.20x = 0.20 m) form a dipole. What is the magnitude of the electric field (in N/C) at the midpoint x=0.10x = 0.10 m?

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4 of 7

Charges q1=+4q_1 = +4 nC (at x=0x = 0) and q2=+1q_2 = +1 nC (at x=0.60x = 0.60 m) are fixed. At what position xx (in m) between them is the electric field zero?

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5 of 7

A dipole consists of q=+5q = +5 nC and βˆ’5-5 nC separated by d=0.040d = 0.040 m. What is the dipole moment (in nCΒ·m)?

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6 of 7

A charge q=βˆ’4 μCq = -4\,\mu\text{C} is placed in a uniform field E=600E = 600 N/C. What is the magnitude of the force (in mN) on the charge?

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7 of 7

The electric field at a distance of r=0.15r = 0.15 m from a point charge is measured to be E=800E = 800 N/C. What is the magnitude of the charge (in nC)?

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Key Takeaways

  • The electric field is a property of space β€” it exists at every point whether or not a test charge is placed there.
  • At any point, Fβƒ—=qEβƒ—\vec{F} = q\vec{E}: positive charges are pushed along Eβƒ—\vec{E}; negative charges are pushed opposite to Eβƒ—\vec{E}.
  • For opposite-sign charge pairs (dipoles), the fields at the midpoint add; for same-sign pairs, they partially cancel.
  • Field lines can never cross β€” if they did, a test charge at that point would have to accelerate in two directions simultaneously.
  • To find a null-field point between two like charges, set the two field magnitudes equal and solve for position.