Notice that now the electric field is perpendicular to the x=axis. It has magnitude

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1 home Physics 415: Lecture 3 Michael Fowler, UVa, 8/9/09 The Dipole Suppose now that in the previous example we replace the lower charge by Q: Q d x-axis -Q y-axis x r E total E = kqrˆ r upper charge Notice that now the electric field is perpendicular to the x=axis. It has magnitude kq kqd kqd E = sin = =. r r 3 x d ( + ) 3/ On the x-axis, this force is maximum at the origin, the point midway between the two charges, and at large distances it decreases as r -3, because the inverse square repulsion and attraction are opposing each other. A pair of equal and opposite charges close together like this is called a dipole. The electric field lines radiate outwards from the positive charge, inwards to the negative charge, and must cross the x-axis at right angles, giving the following picture:

2 Electric field lines of force for a dipole: charges Q, -Q a fixed distance apart Force on a Dipole in an External Electric Field Suppose an electric dipole (imagine now that the two charges are connected by a light insulating rod) is placed in an electric field of uniform strength, so its positive and negative charges feel forces QE which are F = QE F = QE equal in strength, but of course opposite in direction. There is then no net force on the dipole, but there is in general a torque, τ = p E, where the vector dipole moment p has magnitude Q, and direction along the line of the dipole from the negative charge to the positive charge.

3 3 Electric Field from a Continuous Charge Distribution So far, we ve looked at fields from a few charges, apart from the field along the axis from a uniform ring of charge. What if the charge is only on part of the ring? Let s find the field at the center from a uniformly charged arc of a circle, radius, with charge density λ coulombs/meter from 0 to +. 0 The strategy is to find the electric field from one small Arc of circle charged from Ѳ 0 to +Ѳ 0 length of the arc, then add all the small arcs in other words, do an integral. dѳ The length of arc between and + d has length d Ѳ and hence charge λd, so it gives a contribution to the E Ѳ 0 electric field at the center of the circle of magnitude from charge kλd / = kλd / at an angle to the x-axis. in dѳ From the symmetry of the problem, the total field must be along the axis, so we only need count the component kλ/ cosd in that direction: evidently the total field from the whole arc has magnitude ( ) and points in the negative x-direction. ( λ/ ) sin0 E = k It s worth noticing that this field only goes down as 1/, not 1/ : if the radius is doubled, the distance is doubled but so is the amount of charge. We ll see this 1/ behavior in the next paragraph for an infinite line of charge, for a similar reason. Field from a Uniformly Charged Infinite Line This is a slightly more complicated version of the problem above: now r varies, and goes from π / to π /. Again, from the symmetry, the total field must be directly outwards from the infinite line of charge, so we need only calculate the component in that direction. Field from infinite line of charge E from charge in d(tanѳ) secѳ Ѳ tanѳ For uniform line density of charge λ coulombs/meter, the amount of charge corresponding to a small angle d as shown in the diagram is λdy = λd ( tan) = λsec d. This is at a distance sec from the origin (the point where we re finding the field), so contributes an electric field there of strength kq/ r = kλsec d / sec = kλd /. This must be multiplied by cos to give the x-direction component, then integrated from π /to π / to give the total field:

4 4 π / kλ kλ E = cosd = π / And of course it points directly away from the (positively charged) wire. If the wire is not infinite, the limits on the above integral are changed appropriately, and there is also an electric field component parallel to the wire, except at a point level with the center. This component can be found by integrating in the same way. Try drawing the lines of force for this case. Field from a Uniformly Charged Infinite Plane Think of the infinite plane of charge as made up of a huge number of lines of charge parallel to the y- axis, each having charge density λ coulombs per meter, and along the x-axis there is a line density of µ of these lines per meter. This means that there is a area density of charge in the plane of σ = λµ coulombs per square meter. Now, imagine that in the diagram above we used to calculate the field from an infinite line of charge, each bit of the charge line now represents lines of charge perpendicular to the paper this is how we replace the line of charge by a sheet of charge. ecall the electric field strength from charge q was kq / r, that from a line is kλ / r. To find the field from the lines in the angle d, we now use the line density µ in place of λ, so the number of lines in d is µ dy = µ d ( tan ) = µ sec d and their kλ contribution to the electric field is µ sec d. sec The total electric field from the uniformly charged plane is therefore, taking the x-components: π / π / recalling that σ = λµ and k = 1/4 πε 0. kλ σ µ sec cosd = π kλµ = sec ε So the electric field goes out from an infinite uniform sheet of charge perpendicular to it, and does not decrease with distance. For a finite sheet, this picture is still good for distances small compared with the size of the sheet. A common situation (for example, inside a capacitor) is to have two parallel uniform sheets of charge, of equal charge density, but one positive and the other negative. The distance between the sheets is much less than their size, so it s a good approximation to take them as infinite. The electric field is then easily found by superposition: it is a uniform σ / ε0 between the sheets, and zero outside. Electric Fields and Conductors By definition, in a conductor there are charges that are free to move. If there is an electric field, they will move. In electrostatics, all charges are at rest. Therefore, in an electrostatic situation the electric 0

5 field inside every conductor must be zero. It also follows that the lines of force of the electric field outside the conductor must approach the surface perpendicularly, because if there were a parallel component, a current would flow in the surface. 5