Simple Nature - Light and Matter

10.3 Fields by Superposition10.3.1 Electric field of a continuous charge distributionCharge really comes in discrete chunks, but often it is mathematicallyconvenient to treat a set of charges as if they were like acontinuous fluid spread throughout a region of space. For example,a charged metal ball will have charge spread nearly uniformly allover its surface, and for most purposes it will make sense to ignorethe fact that this uniformity is broken at the atomic level. Theelectric field made by such a continuous charge distribution is thesum of the fields created by every part of it. If we let the “parts”become infinitesimally small, we have a sum of an infinitely manyinfinitesimal numbers: an integral. If it was a discrete sum, as inexample 3 on page 564, we would have a total electric field in the xdirection that was the sum of all the x components of the individualfields, and similarly we’d have sums for the y and z components. Inthe continuous case, we have three integrals. Let’s keep it simple bystarting with a one-dimensional example.Field of a uniformly charged rod example 9⊲ A rod of length L has charge Q spread uniformly along it. Findthe electric field at a point a distance d from the center of the rod,along the rod’s axis.⊲ This is a one-dimensional situation, so we really only need todo a single integral representing the total field along the axis. Weimagine breaking the rod down into short pieces of length dz,each with charge dq. Since charge is uniformly spread along therod, we have dq = λ dz, where λ = Q/L (Greek lambda) is thecharge per unit length, in units of coulombs per meter. Sincethe pieces are infinitesimally short, we can treat them as pointcharges and use the expression k dq/r 2 for their contributions tothe field, where r = d − z is the distance from the charge at z tothe point in which we are interested.∫ k dqE z ==r 2∫ +L/2= kλkλ dz−L/2 r 2∫ +L/2−L/2dz(d − z) 2The integral can be looked up in a table, or reduced to an elementaryform by substituting a new variable for d − z. The resultis( ) 1 +L/2E z = kλd − z= kQ L(−L/21d − L/2 − 1d + L/2).a / Example 9.Section 10.3 Fields by Superposition 573