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By the end of this section, you will be able to:
  • Explain what a continuous source charge distribution is and how it is related to the concept of quantization of charge
  • Describe line charges, surface charges, and volume charges
  • Calculate the field of a continuous source charge distribution of either sign

The charge distributions we have seen so far have been discrete: made up of individual point particles. This is in contrast with a continuous charge distribution    , which has at least one nonzero dimension. If a charge distribution is continuous rather than discrete, we can generalize the definition of the electric field. We simply divide the charge into infinitesimal pieces and treat each piece as a point charge.

Note that because charge is quantized, there is no such thing as a “truly” continuous charge distribution. However, in most practical cases, the total charge creating the field involves such a huge number of discrete charges that we can safely ignore the discrete nature of the charge and consider it to be continuous. This is exactly the kind of approximation we make when we deal with a bucket of water as a continuous fluid, rather than a collection of H 2 O molecules.

Our first step is to define a charge density for a charge distribution along a line, across a surface, or within a volume, as shown in [link] .

Figure a shows a long rod with linear charge density lambda. A small segment of the rod is shaded and labeled d l. Figure b shows a surface with surface charge density sigma. A small area within the surface is shaded and labeled d A. Figure c shows a volume with volume charge density rho. A small volume within it is shaded and labeled d V. Figure d shows a surface with two regions shaded and labeled q 1 and q2. A point P is identified above (not on) the surface. A thin line indicates the distance from each of the shaded regions. The vectors E 1 and E 2 are drawn at point P and point away from the respective shaded region. E net is the vector sum of E 1 and E 2. In this case, it points up, away from the surface.
The configuration of charge differential elements for a (a) line charge, (b) sheet of charge, and (c) a volume of charge. Also note that (d) some of the components of the total electric field cancel out, with the remainder resulting in a net electric field.

Definitions of charge density:

  • λ charge per unit length ( linear charge density    ); units are coulombs per meter (C/m)
  • σ charge per unit area ( surface charge density    ); units are coulombs per square meter ( C / m 2 )
  • ρ charge per unit volume ( volume charge density    ); units are coulombs per cubic meter ( C / m 3 )

Then, for a line charge, a surface charge, and a volume charge, the summation in [link] becomes an integral and q i is replaced by d q = λ d l , σ d A , or ρ d V , respectively:

Point charge: E ( P ) = 1 4 π ε 0 i = 1 N ( q i r 2 ) r ^
Line charge: E ( P ) = 1 4 π ε 0 line ( λ d l r 2 ) r ^
Surface charge: E ( P ) = 1 4 π ε 0 surface ( σ d A r 2 ) r ^
Volume charge: E ( P ) = 1 4 π ε 0 volume ( ρ d V r 2 ) r ^

The integrals are generalizations of the expression for the field of a point charge. They implicitly include and assume the principle of superposition. The “trick” to using them is almost always in coming up with correct expressions for dl , dA , or dV , as the case may be, expressed in terms of r , and also expressing the charge density function appropriately. It may be constant; it might be dependent on location.

Note carefully the meaning of r in these equations: It is the distance from the charge element ( q i , λ d l , σ d A , ρ d V ) to the location of interest, P ( x , y , z ) (the point in space where you want to determine the field). However, don’t confuse this with the meaning of r ^ ; we are using it and the vector notation E to write three integrals at once. That is, [link] is actually

E x ( P ) = 1 4 π ε 0 line ( λ d l r 2 ) x , E y ( P ) = 1 4 π ε 0 line ( λ d l r 2 ) y , E z ( P ) = 1 4 π ε 0 line ( λ d l r 2 ) z .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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