Let\'s solve this problem from the electrostatics assignment again. The bar here
ID: 1588140 • Letter: L
Question
Let's solve this problem from the electrostatics assignment again. The bar here has lambda = lambda_o. Note the distance D extends from P to the point exactly halfway across the charge distribution. Determine the electric potential at point P. Determine the electric field at point P using your result from a. Note that D is a coordinate variable here and that Dcap = -xcap so your answer will "appear" to be the negative of what we got previously. Determine V(r) everywhere in space for a sphere of radius R with a constant charge density of rho = rho_o. To do this, you will first need to use Gauss' Law to find the electric field everywhere in space. Then you must use the integral form of the connection between electric potential and electric field: deltaV = -integral E vector.ds vector. Since we want the electric potential to be equal to zero at r = infinity, we should begin this integral at infinity. So the general expression becomes deltaV = V(r) - V(infinity) = - integral^r_infinityE vector dr vector (where the math should make it clear that V(infinity) = 0). Since the electric field may change as a function of radius, this integral may need to be broken apart. Dust grains (r ~+ 1 mum) in Saturn's rings are in a region of dilute ionized gas. Thus, these dust grains can pick up electrons. If the electric potential at the surface of a dust grain is -400 V, determine the number of electrons on the surface of that dust grain. Assume that the electrons are distributed spherically and that the individual dust grains are far enough apart that their electrons don't affect each other.Explanation / Answer
potential on spherical surface = kQ/r
so, 400 = kQ/r
400 = 9*10^9*Q/1*10^-6
Q = 4.44*10^-14C
now Q = ne (here n = no. of electron, e = 1.6*10^-19C)
so, 4.44*10-14 = n*1.6*10^-19
no. of electron = 2.77*10^5
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