Download Electrostatics II-Classical Physics-Handouts and more Lecture notes Classical Physics in PDF only on Docsity! PHYSICS –PHY101 VU © Copyright Virtual University of Pakistan 64 cosdE θ dE θ θ R x y z dsλ P Summary of Lecture 23 – ELECTROSTATICS II 1. In the last lecture we learned how to calculate the electric field if there are any number of point charges. But how to calculate this when charges are continuously distributed over some reg 1 2 3 ion of space? For this, we need to break up the region into little pieces so that each piece is small enough to be like a point charge. So, , or is the total eli E E E E E E = Δ + Δ + Δ + ⋅ ⋅ ⋅⋅ = Δ∑ ectric field. Remember that is a vector that can be resolved ˆˆ ˆ into components, . In the limit where the pieces are small enough, we can write it as an integral, (or x y z E E E i E j E k E dE = + + = ∫ , , ) 2. Charge Density: when the charges are continuously distributed over a region - a line, the surface of a material, or inside a sphere - we must specify the x x y y z zE dE E dE E dE charge de = = =∫ ∫ ∫ . Depending upon how many dimensions the region has, we define: (a) For linear charge distribution: (b) For surface charge distribution: nsity dq ds dq dA λ σ = = (c) For volume charge distribution: The dimensions of , , are determined from the above definitions. 3. As an example of how we work out the electric field coming from a continuous dq dVρ λ σ ρ = charge distribution, let us work out the electric field from a uniform ring of charge at the point P. ( )2 2 20 0 The small amount of charge gives rise to an electric field whose magnitude is 1 4 4 The component in the z direction is cos z ds ds dsdE r z R dE dE λ λ λ πε πε θ = = + = ( )1/ 22 2 with cos .z z r z R θ = = + docsity.com PHYSICS –PHY101 VU © Copyright Virtual University of Pakistan 65 ( ) ( ) ( ) ( ) ( ) 3 / 22 2 0 3 / 2 3 / 2 3 / 22 2 2 2 2 2 0 0 0 So . Since , which is the arc length, does not depend upon or , 4 2 . Answer!! 4 4 4 Note that if you are very far away, t z z z dsdE s z R z R z Rz qzE ds z R z R z R λ πε λ πλ πε πε πε = + = = = + + + ∫ ( )2 0 1he ring looks like a point: , . 4 4. As another example, consider a continuous distribution of charges along a wire that lies along the -axis, as shown below. We want to know the ele z qE z R z z πε = >> ctric field at a distance from the wire. By symmetry, the only non-cancelling component lies along the -axis. x y 2 2 2 0 0 Applying Coulomb's law to the small amount of charge along the axis gives, 1 1 4 4 the component along the direction is cos .y dz z dq dzdE r y z y dE dE λ λ πε πε θ = = + = 2 2 00 0 2 Integrating this gives, cos 2 cos cos . 2 The rest is just technical: to solve the integral, put tan sec . And so, z z z y z z z dzE dE dE dE y z z y dz y d E λθ θ θ πε θ θ θ λ =∞ =∞ =∞ =−∞ = = = = = = + = ⇒ = = ∫ ∫ ∫ ∫ / 2 0 00 cos . Now, we could have equally well taken the x axis. The 2 2 only thing that matters is the distance from the wire, and so the answer is better written as: d y y θ π θ λθ θ πε πε = = =∫ 0 . 2 5. The of any vector field is a particularly important concept. It is the measure of the "flow" or penetration of the field vectors thro E r λ πε = flux ugh an imaginary fixed surface. So, if there is a uniform electric field that is normal to a surface of area , the flux is . More generally, for any surface, we divide the surface up into A EAΦ = little pieces and take the zdE ydE dE yy r θ θ P dqdz z z x docsity.com
