Important Points from Lecturesa) Positive and Negative Charge
attraction and repulsion
(+) --> <-- (-), <-- (+) (+) -->, <-- (-) (-) -->
Charged rod touches neutral electroscope. Electroscope becomes charged. Charge on electroscope redistributes. "Leaves" have same charge, so they repel each other. At some separation, electrostatic force equals gravity so it comes to rest. Height increases with strength of force so it is used to measure charge.
c) Conservation of charge
d) Coulomb's Law
magnitude of force between two charges. Direction of the force is along line joining charges, in directions given by a (above).
Force on a charge due to a collection of other charges is the sum of the forces due to every individual other charge considered separately.
f) Electric Field
g) Gauss' Law
Flux through a closed surface is proportional to the charge contained within the surface.
No field inside a conductor, field at surface always perpendicular to surface.
i) Field lines
Point in direction of field, density indicates strength
j) Electric potential
Work required to move a charge from one point to another is equal to the increase in potential energy. Potential (φ) is energy per unit charge.
Sample Problems1) What is the magnitude of the force a 30µC charge experiences when it is 40cm away from a 3mC charge?
- F=kQ1Q2/r2 = 5000N
First, let's find the field at the top left corner - we'll call this point p.
Use superposition to find the fields at p due to the other three corners (let's call them 1, 2 and 3 respectively clockwise around the square from point p) separately:1) |E1|=kQ/12. As Q1 is a positive charge, the field must point away from it - in the negative x direction.
2) |E2|=kQ/(√2)2. Q2 is negative, so the field points towards it - this vector is 45o below the x-axis: (1/√2, -1/√2)
3) |E3|=kQ/12. As Q3 is a positive charge, the field must point away from it - in the positive y direction.
Applying superposition, the total field is:
- E = kQ*[(-1,0)+(0,1)+1/2(1/√2,-1/√2)]=kQ*(1/2√2-1,1-1/2√2)
- F = -QE = -kQ2*(1/2√2-1,1-1/2√2) = -kQ2(2√2-1)/(2√2)*(-1,1)
What about the force on the other charges? Symmetry tells us immediately that each of them has the exact same force (magnitude) all directed towards the center.
More Gauss' Law ExamplesWhat is the electric field near an infinitely long uniform line of charge?
- 1) Let's define λ as the charge per unit length (units: coulomb/meter).
2) Symmetry tells us that the field must everywhere be radiating directly outwards from the line and have the same magnitude at all points at a given distance from the line. (This can be shown by considering a rotation of the line about its axis and the mirror inversion of the wire along its length)
3) To use Gauss' Law, we need to construct a surface- the obvious choice is a cylinder L meters long with a radius r, centered on the line.
Then, the field is the same magnitude every where on the curved face and also perpendicular to the face, but parallel to the end faces.
- 2πrLE(r) = λL/ε0
E(r) = λ/2πrε0
Construct a "pillbox" for surface:
Horizontal cross section A, extends height h above and below plane.
Flux through surface: 2AE(h)=Q(A)/ε0=Aλ/ε0