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An in-depth analysis of electric field lines and potential generated by point charges and line charges. It covers the calculation of electric fields using coulomb's law, the concept of superposition of potentials, and the relationship between electric potential and electric fields. Students will learn how to find electric potential for various charge distributions using integrals.
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LECTURE #2 Part B
Electric Field Lines Due To A Point Charge Q
When working with point charges, and the electric field or forces exerted on the charges, you may be required to perform vector analysis to define the unit vector ˆ r of the separation between the charges.
LINE OF CHARGES
Shown below is a line of charge. The line has a length of 2a and a linear charge density ρL [C/m]. The elemental length of charge is defined as dz. So what's the electric field at point P a distant r from the line?
Line Of Charge Distribution
The incremental electric field at point P from incremental line of charge dz is:
L o
dz dE r z
ρ π ε
Note that the equation is simply a modified application of Coulomb's law where ρL is equivalent to Q / length and the radial distance between the elemental line of charge dz and point P is
r 2 + z^2.
The incremental electric field component perpendicular to the line of charge is then:
r 2 2
r dE dEcos dE r z
= θ =
Now, since the point P that we are interested in is located such that one half (top) of the line of charge cancels the effects of the other half (bottom) of the line of charge, the components add to zero. The radial field at point P is then the integral of dEr over the total length of charge 2a:
L a L r (^) a o o
r dz E r z r r a
ρ ρ π ε πε
−
If the length of the line charge is much greater than the distance to the point P, then
L r o
r
ρ πε
= [V/m].
Suppose we move a charge along a line from point a to point b. The electric potential difference ∆V is then the work per unit charge in joules per coulomb. We EEs know this unit as Volts. So equation-wise, the electric potential difference is simply: ∆V = E ∆x. Or alternately, the electric field is ∆E = ∆V / ∆x.
Alright, but there is some directionality associated with voltages so in terms of vectors,
ˆ ˆ^ V E x
E x x.
What the heck is the negative sign mean? Well, it reminds us that moving against an electric field means exerting positive work to move the point charge. Ok, so what is the voltage difference when moving a charge from point a to b?
b b a a
So wait! We can generalize and say that