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Unformatted text preview: Solution Derivations for Capa #6 1) Charges are distributed with uniform charge density = 4 . 35 C/m along a semicircle of radius R = 26 . cm centered at the origin of a coordinate system (as shown in the diagram below.) What is the potential at the origin? = Given r = Given The potential difference between two points is defined as V = Z B A E dl The first thing to remember is that potential is a scalar. Thus, the direction of E is irrelevant. The length dl away from the charged surface is a constant r . In my case, it was antiparallel to E (the ring is positively charged). This means that the dot product is negative. If E points in the same direction as r , then the dot product will be positive. It will also be easier to switch to polar coordinates when integrating. Thus, V = Z E dl = Z  rE = Z rE = Z r k dq r 2 = Z k r dq where the last two steps follow because the electric field at each portion of the arc can be regarded as that of a point charge. Since r and k are constants, they can be pulled out of the integral. Thus, V = k r Z dq The limits of integration are in terms of , but the differential is in terms of charge. We must relate the two. dq can be found from , which is charge per length. Thus, a charge dq equals a length times . This length is a portion of the arc over which we are integrating. Thus, dq = r d V = k r Z r d = r k r Z d = k (  0) = k 2) An isolated metal sphere of radius 10 . 5 cm is at a potential of 5200 V . What is the charge on the sphere? r = Given V = Given 1 The potential for a point charge (which essentially what the sphere reduces to if one is outside the sphere) is V = kq r Thus, q = rV k 3) Determine the energy density of the electric field outside the sphere and integrate this throughout all space in order to calculate the total energy in the electric field. The energy density of an electric field is u E = 1 2 E 2 which is on page 647 of the textbook. What this question is asking is to find the energy density associated with the electric field in space and add it all up. Thus, you would have to start at all points right outside the spheres radius, take every point in 3space, and do a calculation for it. Might as well get started; it will take a while. Actually there is a shortcut way. An integral will do this summation for you. But instead of using the radial vector r from the electric field and translating its coordinates into rectangular coordinates, we can transform our variables into spherical coordinates and integrate over all of their range. Since there are three variables in 3space, we will be performing a triple integral. Without using much Calculus 3, the proof for this integral is a little messy. You can look it up in the calculus 3 portion of your calc book. I am not sure how it is presented, but I have a different calc book and this is the way it was presented there....
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 Fall '08
 ROGERS
 Physics, Charge

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