Sheet 07 Uniform Charge Distribution and Application of Gauss’s Theorem.
A plastic rod of length \(2.2m\) and radius \(3.6mm\) carries a negative charge of \(\left(3.8\times10^{-7}\right)C\) spread uniformly over its surface. What is the electric field near the mid-point of the rod, at a point on its surface.
Ans: \(-\left(8.6\times10^{5}\right)\frac{N}{C}\)
A uniformly charged conducting sphere of radius \(1.2m\) has a surface charge density of \(16\frac{\mu C}{m^{2}}\). Find: (i) The charge on the sphere? and (ii) The total electric flux leaving the surface of the sphere?
Ans:(i) \(q=\left(2.9\times10^{-4}\right)C\) and (ii) \(\phi_{E}=\left(3.28\times10^{7}\right)\frac{Nm^{2}}{C}\).
A particle of mass \(\left(5\times10^{-6}\right)g\) is kept over a large horizontal sheet of charge density \(\left(4\times10^{-6}\right)\frac{C}{m^{2}}\). What charge should be given to this particle so that if released it does not fall down under the action of gravity? How many electrons should be removed to give this charge?
Ans: \(q=\left(2.17\times10^{-13}\right)C\) and \(n=\left(1.36\times10^{6}\right)\).
A wire AB of length \(L\) has linear charge density of \(\lambda=kx\), where \(x\) is measured from the end A of the wire. This wire is enclosed by a gaussian hollow surface. Find: the expression for the electric flux through the surface.
Ans: \(\phi_{E}=\frac{KL^{2}}{2\epsilon_{o}}\).
A large plane sheet of charge having surface charge density \(\left(5.0\times10^{-16}\right)\frac{C}{m^{2}}\) lies in the X_Y plane. Find: the electric flux through a circular area of radius \(0.1m\), if the normal to the circular area makes an angle of \(60^{o}\) with the Z-axis. Given: \(\epsilon_{o}=\left(8.85\times10^{-12}\right)\frac{C^{2}}{Nm^{2}}\).
