A charge of 18 nC is uniformly distributed along a straight rod of length 3.0 m

Question

A charge of 18 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius of 2.5 m. What is the magnitude of the electric field at the center of curvature of the arc? please show work

Explanation / Answer

2. Relevant equations E=KQ/R^2 3. The attempt at a solution dQ= ? ds ? =Q/ p R dQ= ? Rd T magnitude dE=kdQ/R^2=K ? Rd T /R^2=KQd T /R^2 sin T magnitude dE=KQd T sin T /R^2 magnitude Ey=KQ[-cos T ]from a to b/ p R^2 the limits of integration would be a=0 and b=4.7(2 ? )/2 ? R I got E=4.31e-1 n/c but it was wrong. PhysOrg.com science news on PhysOrg.com >> Scientists trick iron-eating bacteria into breathing electrons instead >> WhatsApp messaging breached privacy laws >> Crucial, long-overdue BlackBerry makeover arrives Jun22-08, 09:29 PM #2 dynamicsolo Recognitions: Gold Member Homework Helper EDIT: On looking through all of what you wrote, I can tell you this. It is not correct that ?=Q/pR , but you can just use dQ=?Rd? . However, in your field calculation, look at what field component cancels out in the integration. (You won't integrate dE sin ? ...). Make a picture of the arc and the point at its center. Choose an axis through the center that is symmetrically placed through the arc; this will let you take advantage of a symmetry consideration and also clarify how to set up the field integration. Jun23-08, 04:00 PM #3 brett812718 thanks Jul3-08, 05:50 PM Re: electric field of a charged arc #4 SamTaylor Why is dQ=?Rd? ? When ? has the Unit [?]=Cm For a circle I understand it. dQ=Q(dl2pr)=?dl Jul3-08, 11:14 PM Re: electric field of a charged arc #5 Gear300 Make sure to take into account that the electric field at a point is a vector. Jul4-08, 04:13 AM Re: electric field of a charged arc #6 SamTaylor Don't I do that with dEx=dEcos(?) E=Ex=?-??dEcos(?) I thought dQ=?Rd? is the charge density, which is a scalar. Jul4-08, 04:48 AM Re: electric field of a charged arc #7 cryptoguy Quote by SamTaylor Why is dQ=?Rd? ? When ? has the Unit [?]=Cm For a circle I understand it. dQ=Q(dl2pr)=?dl You need to express dQ as a function of d(theta) so you can integrate from 0 to 2*(pi) (if it's a circle) ?=18nC4.7m Jul4-08, 09:18 AM Re: electric field of a charged arc #8 SamTaylor Quote by cryptoguy You need to express dQ as a function of d(theta) so you can integrate from 0 to 2*(pi) (if it's a circle) ?=18nC4.7m I think I did not express myself the right way. Sorry The only thing I don't understand is why there is a R inside dQ=?Rd? . The first time i tried to solve it I used dQ=?d? Because as you said, I need a function of theta. This is how it worked for the circle with dQ=?dl So for the arc it is [?]=C°m I can't interpret that geometrically, it seems to me just to make it fit right Jul4-08, 09:32 AM Re: electric field of a charged arc #9 alphysicist Recognitions: Homework Helper Hi SamTaylor, Quote by SamTaylor I think I did not express myself the right way. Sorry The only thing I don't understand is why there is a R inside dQ=?Rd? . The first time i tried to solve it I used dQ=?d? Because as you said, I need a function of theta. This is how it worked for the circle with dQ=?dl So for the arc it is [?]=C°m I can't interpret that geometrically, it seems to me just to make it fit right As a first look, we know it can't be dQ=?d? because that doesn't have the right units--coulombs on the left, and (coulombs/meter) on the right. The linear charge density ? is the charge per length, and here the length is along the arc. So if s is the length along the arc, a true statement to begin with would be dQ=?ds

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