- 1 -Chapter 2. Electrostatics2.1. The Electrostatic FieldTo calculate the force exerted by some electric charges, q1, q2, q3, ... (the source charges) onanother charge Q (the test charge) we can use the principle of superposition. This principlestates that the interaction between any two charges is completely unaffected by the presence ofother charges. The force exerted on Q by q1, q2, and q3 (see Figure 2.1) is therefore equal to thevector sum of the force F 1 exerted by q1 on Q, the force F 2 exerted by q2 on Q, and the force F 3exerted by q3 on Q.q3q1q2QF2F3F1FtotFigure 2.1. Superposition of forces.The force exerted by a charged particle on another charged particle depends on theirseparation distance, on their velocities and on their accelerations. In this Chapter we willconsider the special case in which the source charges are stationary.The electric field produced by stationary source charges is called and electrostatic field.The electric field at a particular point is a vector whose magnitude is proportional to the totalforce acting on a test charge located at that point, and whose direction is equal to the direction of- 2 -the force acting on a positive test charge. The electric field E , generated by a collection ofsource charges, is defined asE =F Qwhere F is the total electric force exerted by the source charges on the test charge Q. It isassumed that the test charge Q is small and therefore does not change the distribution of thesource charges. The total force exerted by the source charges on the test charge is equal toF = F 1+ F 2+ F 3+ ... =14pe0q1Qr12ˆ r 1+q2Qr22ˆ r 2+q3Qr32ˆ r 3+ ...Ê Ë Á ˆ ¯ ˜ =Q4pe0qiri2ˆ r ii =1nÂThe electric field generated by the source charges is thus equal toE =F Q=14pe0qiri2ˆ r ii =1nÂIn most applications the source charges are not discrete, but are distributed continuously oversome region. The following three different distributions will be used in this course:1. line charge l: the charge per unit length.2. surface charge s: the charge per unit area.3. volume charge r: the charge per unit volume.To calculate the electric field at a point P generated by these charge distributions we have toreplace the summation over the discrete charges with an integration over the continuous chargedistribution:1. for a line charge: E P ()=14pe0ˆ r r2ldlLineÚ2. for a surface charge: E P ()=14pe0ˆ r r2sdaSurfaceÚ3. for a volume charge: E P ()=14pe0ˆ r r2rdtVolumeÚ- 3 -Here ˆ r is the unit vector from a segment of the charge distribution to the point P at which weare evaluating the electric field, and r is the distance between this segment and point P .Example: Problem 2.2a) Find the electric field (magnitude and direction) a distance z above the midpoint between twoequal charges q a distance d apart. Check that your result is consistent with what you wouldexpect when z » d.b) Repeat part a), only this time make he right-hand charge -q instead of +q.zd/2d/2Pa)ErElEtotzd/2d/2Pb)ElErEtotFigure 2.2. Problem 2.2a) Figure 2.2a shows that the x components of the electric fields generated by the two pointcharges cancel. The total electric field at P is equal to the sum of the z components of theelectric fields generated by the two point charges:E P ()= 214pe0q14d2+ z2Ê Ë ˆ ¯ z14d2+ z2ˆ z =12pe0qz14d2+ z2Ê Ë ˆ ¯ 3/2ˆ z When z » d this equation becomes approximately equal toE P ()@12pe0qz2ˆ z =14pe02qz2ˆ z- 4 -which is the Coulomb field generated by a point charge with charge 2q.b) For the electric fields generated by the point charges of the charge distribution shown inFigure 2.2b the z components cancel. The net electric field is therefore equal toE P ()= 214pe0q14d2+ z2Ê Ë ˆ ¯ d214d2+ z2ˆ x =14pe0qd14d2+ z2Ê Ë ˆ ¯ 3/2ˆ x Example: Problem 2.5Find the electric field a distance z above the center of a circular loop of radius r which carriesa uniform line charge l.rzPdEldEr2dEzFigure 2.3. Problem 2.5.Each segment of the loop is located at the same distance from P (see Figure 2.3). Themagnitude of the electric field at P due to a segment of the ring of length dl is equal todE =14pe0ldlr2+ z2- 5 -When we integrate over the whole ring, the horizontal components of the electric field cancel.We therefore only need to consider the vertical component of the electric field generated by eachsegment:dEz=zr2+ z2dE =ldl4pe0zr2+ z2()3/2The total electric field generated by the ring can be obtained by integrating dEz over the wholering:E =l4pe0zr2+ z2()3/2dlRingÚ=14pe0zr2+ z2()3/22pr()l=14pe0zr2+ z2()3/2qExample: Problem 2.7Find the electric field a distance z from the center of a spherical surface of radius R, whichcarries a uniform surface charge density s. Treat the case z < R (inside) as well as z > R(outside). Express your answer in terms of the total charge q on the surface.Pzrcosqrsinqz-rcosqqFigure 2.4. Problem 2.7.Consider a slice of the shell centered on the z axis (see Figure 2.4). The polar angle of thisslice is q and its width is dq. The area dA of this ring isdA = 2pr sinq()rdq= 2pr2sinqdqThe total charge on this ring is equal to- 6 -dq =sdA =12qsinqdqwhere q is the total charge on the shell. The electric field produced by this ring at P can becalculated using the solution of Problem 2.5:dE =18pe0qrz - r cosqr2+ z2- 2zr cosq()3/2r sinqdqThe total field at P can be found by integrating dE with respect to q:E =18pe0qrz - r cosqr2+ z2- 2zr cosq()3/2r sinqdq0pÚ==18pe0qrz - r cosqr2+ z2- 2zr cosq()3/2drcosq()0pÚ=18pe0qrz - yr2+ z2- 2zy()3/2dy-rrÚThis integral can be solved using the following relation:z - yr2+ z2- 2zy()3/2=-ddz1r2+ z2- 2zySubstituting this expression into the integral we obtain:E =-18pe0qrddz1r2+ z2- 2zydy-rrÚ=-18pe0qrddzr2+ z2- 2zy-z-rr==-18pe0qrddzr + z()- r - zzÏ Ì Ó ¸ ˝ ˛ Outside the shell, z > r and consequently the electric field is equal toE =-18pe0qrddzr + z()- z - r()z=-14pe0qddz1z=14pe0qz2Inside the shell, z < r and consequently the electric field is equal toE =-18pe0qrddzr + z()- r - z()z=-14pe0qrddz1 = 0- 7 -Thus the electric field of a charged shell is zero inside the shell. The electric field outside theshell is equal to the electric field of a point charge located at the center of the shell.2.2. Divergence and Curl of Electrostatic FieldsThe electric field can be graphically represented using field lines. The direction of the fieldlines indicates the direction in which a positive test charge moves when placed