May 2011 - Career Point

of the charge distribution. In either case, begin youranalysis by asking the question, "What is thesymmetry ?"Step 2 : Set up the problem using the following stepsSelect the surface that you will use with Gauss'slaw. We often call it a Gaussian surface. If youare trying to find the field at a particular point,then that point must lie on your Gaussian surface.The Gaussian surface does not have to be a realphysical surface, such as a surface of a solidbody. Often the appropriate surface is animaginary geometric surface; it may be in emptyspace, embedded in a solid body, or both.Usually you can evaluate the integral in Gauss'slaw (without using a computer) only if theGaussian surface and the charge distribution havesome symmetry property. If the chargedistribution has cylindrical or sphericalsymmetry, choose the Gaussian surface to be acoaxial cylinder or a concentric sphere,respectively.Step 3 : Execute the solution as follows :Carry out the integral in Eq.ΦE =∫E cosφ dA =∫E dA =∫Er .dAr Q= enclε0(various forms of Gauss's law)This may look like a daunting task, but thesymmetry of the charge distribution and yourcareful choice of a Gaussian surface makes itstraightforward.Often you can think of the closed surface as beingmade up of several separate surfaces, such as theside and ends of a cylinder. The integral∫ E dAover the entire closed surface is always equal tothe sum of the integrals over all the separatesurfaces. Some of these integrals may be zero, asin points 4 and 5 below.If E r is perpendicular (normal) at every point to asurface with area A, if points outward from theinterior of the surface, and if it equal to EA. Ifinstead E r is perpendicular and inward, then E ⊥ =– E and∫ E ⊥dA = – EA.If E r is tangent to a surface at every point, then E ⊥= 0 and the integral over that surface is zero.If E r = 0 at every point on a surface, the integralis zero.In the integral∫ E ⊥dA , E ⊥ is always theperpendicular component of the total electric fieldat each point on the closed Gaussian surface. Ingeneral, this field may be caused partly bycharges within the surface and partly by chargesoutside it. Even when there is no charge withinthe surface, the field at points on the Gaussiansurface is not necessarily zero. In that case,however, the integral over the Gaussian surface –is always zero.Once you have evaluated the integral, use eq. tosolve for your target variable.Step 4 : Evaluate your answer : Often your result willbe a function that describes how the magnitude of theelectric field varies with position. Examine this functionwith a critical eye to see whether it make sense.1. Supposing that the earth has a charge surface densityof 1 electron/metre 2 , calculate (i) earth's potential, (ii)electric field just outside earths surface. Theelectronic charge is – 1.6 × 10 –19 coulomb and earth'sradius is 6.4×10 6 metre (ε 0 = 8.9 × 10 –12 coul 2 /nt–m 2 ).Sol. Let R and σ be the radius and charge surface densityof earth respectively. The total charge, q on the earthsurface is given byq = 4 p R 2 σ(i) The potential V at a point on earth's surface is sameas if the entire charge q were concentrated at itscentre. Thus,1 qV = .4πε 0 R1 4πR2 σ R. σ= . =4πε0R ε0Substituting the given values−6−192(6.4×10 metre) × ( −1.6×10 coul / metre )V =−122 2(8.9×10 coul / nt − m )(ii) E =Solved Examplesnt − m= – 0.115 coul1 q 12 =4πε0R 4πε . 4πRσ20 R−19joule= – 0.115 = – 0.115 volt.coul−1.6×10 coul / metre== – 1.8 × 10 –8 nt/coul.−122 28.9×10 coul / nt − mThe negative sign shows that E is radially inward.2. Determine the electric field strength vector if thepotential of this field depends on x, y co-ordinates as(a) V = a(x 2 – y 2 ) and (b) V = axy.Sol. (a) V = a(x 2 – y 2 )∂V∂VHence, E x = – = – 2ax, E y = – = + 2ay∂ x ∂ y∴ E = – 2axi + 2ayjor E = – 2a(xi – yj)(b) V = a x y22=σε 0XtraEdge for IIT-JEE 23 MAY 2011