Chapter 2. Electrostatics - University of Rochester

1 - 1 - Chapter 2. 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 ) 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 Superposition of force exerted by a charged particle on another charged particle depends on theirseparation distance, on their velocities and on their accelerations.

2 In this Chapter we willconsider the special case in which the source charges are electric field produced by stationary source charges is called and electrostatic 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.

3 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+..=14pe0q1Qr12 r 1+q2Qr22 r 2+q3Qr32 r 3+.. =Q4pe0qiri2 r ii=1n The electric field generated by the source charges is thus equal toE =F Q=14pe0qiri2 r ii=1n In most applications the source charges are not discrete, but are distributed continuously oversome region. The following three different distributions will be used in this charge l: the charge per unit charge s: the charge per unit charge r: the charge per unit 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.

4 For a line charge: E P ()=14pe0 r r2ldlLine 2. for a surface charge: E P ()=14pe0 r r2sdaSurface 3. for a volume charge: E P ()=14pe0 r r2rdtVolume - 3 -Here 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 ) 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 ) Repeat part a), only this time make he right-hand charge -q instead of + )ErElEtotzd/2d/2Pb)ElErEtotFigure Problem )Figure shows that the x components of the electric fields generated by the two pointcharges cancel.

5 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+z2 z14d2+z2 z =12pe0qz14d2+z2 3/2 z When z d this equation becomes approximately equal toE P ()@12pe0qz2 z =14pe02qz2 z - 4 -which is the Coulomb field generated by a point charge with charge )For the electric fields generated by the point charges of the charge distribution shown inFigure the z components cancel. The net electric field is therefore equal toE P ()=214pe0q14d2+z2 d214d2+z2 x =14pe0qd14d2+z2 3/2 x Example: Problem the electric field a distance z above the center of a circular loop of radius r which carriesa uniform line charge Problem segment of the loop is located at the same distance from P (see Figure ).

6 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 therefore only need to consider the vertical component of the electric field generated by eachsegment:dEz=zr2+z2dE=ldl4pe0zr2+z2() 3/2 The total electric field generated by the ring can be obtained by integrating dEz over the wholering:E=l4pe0zr2+z2()3/2dlRing =14pe0zr2+z2()3/22pr()l=14pe0zr2+z2()3/2 qExample: Problem the electric field a distance z from the center of a spherical surface of radius R, whichcarries a uniform surface charge density s.

7 Treat the case z < R (inside) as well as z > R(outside). Express your answer in terms of the total charge q on the Problem a slice of the shell centered on the z axis (see Figure ). The polar angle of thisslice is q and its width is dq. The area dA of this ring isdA=2prsinq()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 :dE=18pe0qrz-rcosqr2+z2-2zrcosq()3/2rsin qdqThe total field at P can be found by integrating dE with respect to q:E=18pe0qrz-rcosqr2+z2-2zrcosq()3/2rsin qdq0p ==18pe0qrz-rcosqr2+z2-2zrcosq()3/2drcosq ()0p =18pe0qrz-yr2+z2-2zy()3/2dy-rr This integral can be solved using the following relation:z-yr2+z2-2zy()3/2=-ddz1r2+z2-2z ySubstituting this expression into the integral we obtain.

8 E=-18pe0qrddz1r2+z2-2zydy-rr =-18pe0qrddzr2+z2-2zy-z-rr==-18pe0qrddzr +z()-r-zz Outside the shell, z > r and consequently the electric field is equal toE=-18pe0qrddzr+z()-z-r()z=-14pe0qddz1z =14pe0qz2 Inside 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 Divergence and Curl of Electrostatic FieldsThe electric field can be graphically represented using field lines.

9 The direction of the fieldlines indicates the direction in which a positive test charge moves when placed in this field. Thedensity of field lines per unit area is proportional to the strength of the electric field. Field linesoriginate on positive charges and terminate on negative charges. Field lines can never crosssince if this would occur, the direction of the electric field at that particular point would beundefined. Examples of field lines produced by positive point charges are shown in Figure )b)Figure a) Electric field lines generated by a positive point charge with charge q.

10 B)Electric field lines generated by a positive point charge with charge flux of electric field lines through any surface is proportional to the number of field linespassing through that surface. Consider for example a point charge q located at the origin. Theelectric flux FE through a sphere of radius r, centered on the origin, is equal toFE=E da Surface =14pe0qr2 r r2sinqdqdf r ()Surface =qe0 Since the number of field lines generated by the charge q depends only on the magnitude of thecharge, any arbitrarily shaped surface that encloses q will intercept the same number of fieldlines.