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Electric Field - University of Kentucky

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Electric Field Coulomb’s Law: F = kq 1 q 2 /r 2 , repel if like signs, attract if unlike signs. Consider the force on a positive test charge (q = +1) from several other (+ and -) charges, Q j . F 3 F 5 F 4 F 1 F 2 F net All the forces, F j , are proportional to the magnitude (q) of the test charge and also change sign if q changes sign. Therefore, the net force (their vector sum) is also proportional to the magnitude of q and changes sign if q changes sign. Therefore, it is useful to define a vector quantity, the electric field (E) to each point in space, such that the net electric force of any charge (q) placed at that point: F net = qE E = F net /q
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Electric Field

Coulomb’s Law: F = kq1q2/r2, repel if like signs, attract if unlike signs.

Consider the force on a positive test charge (q = +1) from several other (+ and -) charges, Qj.

F3

F5

F4F1F2

Fnet

All the forces, Fj, are proportional to the magnitude (q) of the test charge and also change sign if q changes sign. Therefore, the net force (their vector sum) is also proportional to the magnitude of q and changes sign if q changes sign.

Therefore, it is useful to define a vector quantity, the electric field (E) to each point in space, such that the net electric force of any charge (q) placed at that point:

Fnet = qEE = Fnet/q

Fnet = qEE = Fnet/q

The electric field is defined at every point in space. It tells you the magnitude and direction of the net electric force of any test charge placed at the spot.

E around a single + charge E around a single - charge

E of a pair of equal magnitude but opposite sign charges (dipole)

E of a pair of equal positive charges

E of two conducting plates with equal and opposite sign charges (i.e. a capacitor): the electric field between the plates is uniform and outside the plates E ≈ 0

The electric field is a (vector) quantity that is a function of position (r) that depends on all the positions of all other charges. If a charge q is placed at r, E(r) = F/q, where F is the total electrostatic force.

Similarly, the voltage is a scalar function of position that also depends on the positions of all other charges. If a charge q is placed at r, V(r) = U/q, where U is the electrostaticpotential energy.

Is there a relationship between E(r) and V(r)? For gravity (Ch. 2) we found that Fgrav = -gradient(Ugrav).

The same relation holds for the electrostatic force and potential energy.Divide by q: E = -gradient (V) [The book omits the minus sign.]

i.e. The electric field is large if the voltage is changing quickly with position, and points in the direction of the most rapid decrease in V.

E = -gradient (V)

If positive charge is distributed uniformly on the surface of a sphere, this is what the voltage looks like as a function of distance from center.

The electric field = zero inside the sphere, since V = constant. The electric field is largest right outside the sphere, where V is changing most rapidly. E dies off with distance as V flattens out, going to zero.

Distance from center

Consider a conductor in equilibrium. A conductor is a material in which charges (usually electrons) are free to move in response to an electrostatic force, i.e. an electric field. Therefore, they will move around until they make the electric field = 0.

Rule 1: The electric field within a conductor in equilibrium = 0.

Since E = -gradient (V), this immediately gives:

Rule 2: The voltage inside a conductor in equilibrium = constant. (We say that the conductor is an “equipotential”.)

Imagine that the conductor has some charge on it. Since any non-compensated (i.e. not neutralized) charge must create an electric field next to it, this gives us:

Rule 3: Any charge on a conductor in equilibrium must lie on its surface.

Rule 1: The electric field within a conductor in equilibrium = 0.Rule 2: The voltage inside a conductor in equilibrium = constant.Rule 3: Any charge on a conductor in equilibrium must lie on its surface.

This charge on the surface must distribute itself so that the voltage on the surface is constant; otherwise, charges would move down the voltage gradient. →

Rule 4: The voltage on the surface of a conductor in equilibrium = constant (the same constant as in the interior, from Rule 2).

Finally, since E = -gradient (V) and there is no component of the gradient parallel to the surface, E must be perpendicular to the surface. But it cannot point into the conductor (or charge would flow in. →

Rule 5: The electric field immediately outside a conductor is perpendicular to the surface, pointing out.

Rule 1: The electric field within a conductor in equilibrium = 0.Rule 2: The voltage inside a conductor in equilibrium = constant.Rule 3: Any charge on a conductor in equilibrium must lie on its surface.Rule 4: The voltage on the surface of a conductor in equilibrium = constant.Rule 5: The electric field immediately outside a conductor is perpendicular to the surface, pointing out.

So this picture is appropriate for a charged conducting ball.

Here is the picture for a neutral conducting cylinder placed between two conducting plates, with equal and opposite charges (all extending to ± infinity in the direction perpendicular to the screen).

Electric Discharges

Although most of the atoms in air are electrically neutral (i.e. same number of electrons as protons), a few will be “ionized”, i.e. will have lost one or more electrons, which are still bouncing around nearby. [This ionization is usually the result of the atmosphere being bombarded by particles from space, “cosmic rays” which can knock electrons out of their atoms. Also, radiation from small amounts of radioactive elements nearby may also ionize the air.] If there is an electric field present, it will accelerate the electrons and positive ions, but since the electrons are much lighter, they will accelerate more. If a fast electron hits another atom, it can ionize it, and then the freed electron can hit another atom, etc. This ionization trail is what we observe as a discharge, i.e. a spark. It can occur (in dry air) if the electric field reaches ≈ 3,000,000 V/m (= 30,000 V/cm = 3000 V/mm). [The light of the spark comes from electrons falling back into atoms, which we will discuss in Chapter 13.]

Lightning is a discharge from the large electric fields created by charges which accumulate in clouds; intense up and down drafts within clouds ionize water drops, making the bottom of the cloud negative and top positive. In the discharge, collisions of the electrons and ions cause the surrounding, local air to heat (often to 50,000 degrees) and expand rapidly, even faster than the speed of sound. Thunder is the resulting sonic boom. Also, in the intense heat, some of the molecular oxygen (O2) dissociates and recombines to form ozone (O3), giving rise to the characteristic, “sweet” odor of thunderstorms.

The light travels a million times faster than the sound, which travels 331 m/s ≈ 1/5 mile/second.

For the same voltage difference, electric fields are largest near sharp points. Therefore, lightning rods are used to attract lightning, and keep the discharge away from other structures. (A cable carries the electric charges down into the earth, so that the rod remains neutral.)

Electric Currents

Although “xerography” utilizes static electric charges, most applications of electricity utilize currents, in which charges continuously flow through conducting materials (typically metals or “semiconductor”.) Although the flowing charges are usually electrons, by convention we pretend that there is positive current flow.

In most cases, the effects of negative electrons flowing to the right are the same as an equivalent amount of positive current flowing to the left.

We measure current (I) by the amount of charge flowing through a cross-sectional area of the wire (or device) per second: I = ΔQ / Δt

1 amp = 1 Coulomb/second.

If several devices are connected to each other (with wires – to be discussed later), the same current must flow through each; otherwise, charge would build up in a device, repel more charge from entering and stop the current. Therefore, (at least) two wires must enter each device, one taking current in and the other taking current out, and the current must flow in a loop, making a “circuit”.

When a charge goes around the circuit and ends up where it started, it must have the same electric potential energy as when it started. So the sum of the changes in potential energy (ΔUj) in each device must = 0 (i.e. some positive and some negative):

∑ΔUj = 0

But the voltage change in each device Vj = ΔUj/charge.

Therefore ∑Vj = 0

V1

V2 V3

V4

V5

The rate at which energy changes in each device is the power:

Pj = ΔUj / Δt = Vj ⸱ ΔQ / Δt Pj = Vj ⸱ I

[Pj is in watts (W)1 W = 1 J/s = 1V ⸱ 1 A]

If the device puts energy into the circuit (e.g. a battery, which converts chemical energy into electrical energy), the voltage(and potential energy) will increase in the direction of current flow.

If the voltage decreases in the direction of current flow, the device will be using up electrical energy, turning it into heat (most common), light (e.g. a bulb), sound (e.g. a speaker) or back into chemical energy (when you recharge a battery).

Exercises:14. A car battery is labeled as providing 12 V. Compare the electrostatic potential energy of a positive charge on the battery’s negative terminal with that on its positive terminal.15. When technicians work with static-sensitive electronics, they try to make as much of their environment electrically conducting as possible. Why does this conductivity diminish the threat of static electricity?17. Holding your hand on a static generator (for example, a van de Graff generator) can make your hair

stand up, but only if you are standing on a good electrical insulator. Why is that insulator important?19. The electric field around an electrically charged hairbrush diminishes with distance from that hairbrush. Use Coulomb’s law to explain this decrease in the magnitude of the field.20. Which way does the electric field point around the positive terminal of an alkaline battery?

Problems:7. What force will a 0.01-C charge experience in a 5-N/C electric field pointing upward?

8. A sock with a charge of –0.0005 C is in a 1000-N/C electric field pointing toward the right. What force does the sock experience?

9. A piece of plastic wrap with a charge of 0.00005 C experiences a forward force of 0.0010 N. What is the local electric field?


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