Timedora

1 Electric Charge and Fields

Progresses from electric charge and Coulomb’s law to electric fields, flux, and Gauss’s law, showing how these ideas describe and calculate electric interactions.

Charge and its basic properties

is a property of matter with two forms: positive and negative. Like charges repel, while unlike charges attract. A proton has charge +e+e, and an electron has charge −e-e, where the elementary charge has magnitude approximately 1.602×10−19 C1.602\times10^{-19}\,\text{C}. Charge is quantized, so an object’s net charge is an integer multiple of ee. In ordinary charging, electrons transfer between objects; the total charge of an isolated system remains constant.

These properties set up two central questions: what force do charges exert on one another, and how can their influence be described at locations in space?

Force between charges

For two stationary point charges separated by a distance rr, gives the force magnitude:

F=ke∣q1q2∣r2,ke=14πε0.F=k_e\frac{|q_1q_2|}{r^2},\qquad k_e=\frac{1}{4\pi\varepsilon_0}.

The force grows with the magnitude of either charge and decreases with the square of the separation. It acts along the line joining the charges: like signs repel, and opposite signs attract.

For example, charges of +2.0 μC+2.0\,\mu\text{C} and −3.0 μC-3.0\,\mu\text{C}, separated by 0.20 m0.20\,\text{m}, exert a force with magnitude

F=(8.99×109)(2.0×10−6)(3.0×10−6)(0.20)2≈1.35 N.F=(8.99\times10^9)\frac{(2.0\times10^{-6})(3.0\times10^{-6})}{(0.20)^2}\approx1.35\,\text{N}.

Because the charges have opposite signs, the force is attractive. With several source charges, calculate each force as a vector and add the vectors; adding magnitudes alone would lose directional information.

and superposition

An describes the force a source charge would exert per unit positive test charge. The field and the force on a test charge are related by

E=Fqtest,F=qtestE.\mathbf E=\frac{\mathbf F}{q_{\text{test}}},\qquad \mathbf F=q_{\text{test}}\mathbf E.

The field is measured in newtons per coulomb. Its direction is defined by the force on a positive test charge: field lines point away from positive charges and toward negative charges.

For a point source charge qq, the field at distance rr is

E=keqr2r^,\mathbf E=k_e\frac{q}{r^2}\hat{\mathbf r},

where r^\hat{\mathbf r} points from the source to the field point. A positive source produces an outward field; a negative source produces an inward field.

The says that fields from multiple charges add vectorially. For a continuous charge distribution, the contributions can be summed by integration. For example, a +2.0 nC+2.0\,\text{nC} point charge produces a field of magnitude approximately 2.0×102 N/C2.0\times10^2\,\text{N/C} at a distance of 0.30 m0.30\,\text{m}, directed outward.

Takeaway: calculates forces between charges; the describes the influence of source charges at each point in space.

through surfaces

measures how much field passes through a surface. For a flat surface in a uniform field,

ΦE=E⋅A=EAcos⁡θ,\Phi_E=\mathbf E\cdot\mathbf A=EA\cos\theta,

where AA is the surface area and θ\theta is the angle between the field and the area vector. The area vector is perpendicular to the surface. Flux is greatest when the field points along this vector, so it crosses the surface perpendicularly; flux is zero when the field is parallel to the surface.

For a curved surface or a field that varies across the surface, add the contributions over the surface:

ΦE=∫SE⋅dA.\Phi_E=\int_S\mathbf E\cdot d\mathbf A.

For a closed surface, the area vectors point outward and the integral uses a closed-surface symbol. Flux is measured in N⋅m2/C\text{N}\cdot\text{m}^2/\text{C}.

For example, a uniform field of 400 N/C400\,\text{N/C} crossing a flat surface of area 0.020 m20.020\,\text{m}^2 at an angle of 60∘60^\circ to its area vector gives

ΦE=EAcos⁡θ=4.0 N⋅m2/C.\Phi_E=EA\cos\theta=4.0\,\text{N}\cdot\text{m}^2/\text{C}.

and symmetry

connects the net flux through any closed surface to the net charge enclosed:

∮SE⋅dA=qencε0.\oint_S\mathbf E\cdot d\mathbf A=\frac{q_{\text{enc}}}{\varepsilon_0}.

Only the net enclosed charge determines the total flux. Charges outside the surface may affect the field at points on it, but they do not change the net flux through it.

A is an imaginary closed surface chosen to help analyze a charge distribution. is always valid, but it is especially useful for finding the field when the distribution has enough symmetry—typically spherical, cylindrical, or planar symmetry—to simplify the field over the chosen surface.

For a point charge qq, choose a spherical of radius rr. Symmetry makes the field radial and equal in magnitude everywhere on the sphere, so

E(4πr2)=qε0⟹E=keqr2.E(4\pi r^2)=\frac{q}{\varepsilon_0}\quad\Longrightarrow\quad E=k_e\frac{q}{r^2}.

This recovers the point-charge field. In general, match the to the symmetry, determine the enclosed charge, and use the flux integral to solve for the field.

Takeaway: always relates closed-surface flux to enclosed charge; symmetry often makes it a practical tool for calculating fields.