May 2011 - Career Point

In a conductor, all points have the same potential.If charge q (coulomb) is placed at a point where thepotential is V (volt), the potential energy of thesystem is qV (joule). It follows that if charges q 1 , q 2are separated by distance r, the mutual potentialqenergy of the system is 1 q 24πε r.• Relation Between Field (E) And Potential (V)The negative of the rate of change of potential alonga given direction is equal to the component of thefield that direction.∂VE r = – along r∂ r∂V and E ⊥ = perpendicular to rr ∂θWhen two points have different potentials, an electricfield will exist between them, directed from thehigher to the lower potential.• Lines of ForceA line of force in an electric field is such a curve thatthe tangent to it at any point gives the direction of thefield at that point. Lines of force cannot intersecteach other because it is physically impossible for anelectric field to have two directions simultaneously.• Equipotential SurfacesThe locus of points of equal potential is called anequipotential surface. Equipotential surfaces lie atright angles to the electric field. Like lines of force,they can never intersect.Note: For solving problems involving electrostaticunits, remember the following conversion factors:3 × 10 9 esu of charge = 1 C1 esu of potential = 300 V• Electric FluxThe electric flux over a surface is the product of itssurface area and the normal component of the electricfield strength on that surface. Thus,dϕ = (E cos θ) ds = E n ds = → E . → dsdsOThe total electric flux over a surface is obtained bysumming :→ →→ →ϕ E = ∑ E . ∆ s or∫ E .d sGauss's Theorem The total electric flux across a1closed surface is equal to times the total chargeε0inside the surface.Mathematically ∑ → E . ∆ →s = q/ε 0where q is the total charge enclosed by the surface.ENProblems in electrostatics can be greatly simplifiedby the use of Gaussian surfaces. These are imaginarysurfaces in which the electric intensity is eitherparallel to or perpendicular to the surface everywhere.There are no restrictions in constructing aGaussian surface.The following results follow from Gauss's law1. In a charged conductor, the entire charge residesonly on the outer surface. (It must always beremembered that the electric field is zero inside aconductor.)2. Near a large plane conductor with a chargedensity σ (i.e., charge per unit area), the electricintensity isE = σ/ε 0 along the normal to the plane3. Near an infinite plane sheet of charge with acharge density σ, the electric intensity isE = σ/2ε 0 along the normal to the plane4. The electric intensity at a distance r from the axisof a long cylinder with λ charge per unit length(called the linear density of charge), is1 λ→E = along r2πεr0Problem solving strategy: Coulomb's Law :Step 1 : The relevant concepts : Coulomb's lawcomes into play whenever you need to know theelectric force acting between charged particles.Step 2 : The problem using the following steps :Make a drawing showing the locations of thecharged particles and label each particle with itscharge. This step is particularly important if morethan two charged particles are present.If three or more charges are present and they donot all lie on the same line, set up an xycoordinatesystem.Often you will need to find the electric force onjust one particle. If so, identify that particle.Step 3 : The solution as follows :For each particle that exerts a force on the particleof interest, calculate the magnitude of that force1 | q1q2 |using equation F =24πε0 rSketch the electric force vectors acting on theparticle(s) of interest due to each of the otherparticles (that is, make a free-body diagram).Remember that the force exerted by particle 1 onparticle 2 points from particle 2 toward particle 1if the two charges have opposite signs, but pointsfrom particle 2 directly away from particle 1 if thecharges have the same sign.Calculate the total electric force on the particle(s)of interest. Remember that the electric force, likeany force, is a vector. When the forces acting on acharge are caused by two or more other charges,the total force on the charge is the vector sum ofXtraEdge for IIT-JEE 21 MAY 2011