InterviewSolution
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InterviewSolution
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(a) Using Gauss's law, derive expression for intensity of electric field at any point near the infinitely long straight uniformly charged wire. (b) The electric field components in the following figure are E_(x)=alphax, E_(y)=0, E_(z)=0, in which alpha=400N//Cm. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that a=0.1 m. |
Answer» Solution :(a) CONSIDER an infinitely long straight charged wire of linear charge density `lambda`. To find electric field at a point P situated at a distance r from the wire by using Gauss. law consider a cylinder of length l and radius r as the Gaussian surface.  From SYMMETRY consideration electric field at each point of its curved surface is `vecE` and is pointed outwards normally. Therefore, electric flux over the curved surface `=intvecE.hatn ds=E 2pirl` On the SIDE faces 1 and 2 of the cylinder normal drawn on the surface is perpendicular to electric field E and hence these surfaces do not contribute towards the total electric flux. `therefore `Net electric flux over the entire Gaussian surface `phi_(E)=E.2pirl"....(i)"` By Gauss law electric flux `phi_(E)=(1)/(in_(0))"(charged enclosed)"=(lambdal)/(in_(0))"...(ii)"` Comparing (i) and (ii), we have `E.2pirl =(lambdal)/(in_(0))` `rArr""E=E=(lambda)/(2PI in_(0)r) and vecE=(lambda)/(2pi in_(0)r)hatr` (b) (i) Here, `E_(x)=alpha. x=400x, E_(y)=E_(z)=0 and a=0.1m`. So the electric flux is linked with only two fa c es of the cube lying in Y -Z plane (i.e., perpendicular to `E_(x)`). As position of left face of cube is at `x=a=0.1m`, hence `(E_(x))_(1)=400xx0.1=40NC^(-1)` (inward into the Cube) Now position of right face of cube is `x=a+a=2a=0.2m`, hence `(E_(x))_(2)=400xx0.2=80NC^(-1)`. `therefore"flux "phi_(2)=(E_(x))_(2).A^(2)=80xx0.01=0.8Nm^(2)C^(-1)" (outward)"` `therefore` Net flux `phi_(2)=(E_(x))_(2.)A^(2)=80xx0.01=0.8Nm^(2)C^(-1)" (outward)"` `therefore` Net flux through the cube `phi_(E)=phi_(2)-phi_(1)=0.8-0.4=0.4Nm^(2)C^(-1)` outward (ii) Charge within the cube `q=in_(0).phi_(E)=(8.854xx10^(-12))xx0.4=3.54xx10^(-12)C` |
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