Chapter 22: Electric FieldsProblem 1In Fig. 22-30 the electric field lines on theleft have twice the separation as those on theright. (a) If the magnitude of the field at A is40 N/C, what force acts on a proton at A?(b) What is the magnitude of the field at B?Answer(a) 6.40 ×10-18N. (b) ≈20 N/C.BAFigure 22-30Solution(a)If E is the magnitude of the electric field at A then the magnitude of the force on theproton isF = qE, (q is the magnitude of the charge on proton)F = (1.60 ×10-19C)*(40 N/C)F = 6.40 ×10N(b)Note that the field lines at B are about twice as far apart as at A. Since the density ofthe field lines represents the magnitude of the electric field this means that the magnitude ofthe field at B is about half of that at A or ≈20 N/C.Problem 3What is the magnitude of a point charge whose electric field 50 cm away has the magnitudeof 2.0 N/C?Answer56 pCThe magnitude of the electric field due to a point charge q at a point at a distance r from thepoint charge isE=14πεoqr2Therefore the magnitude of the charge isq=4πεor2E=0.50m( )22.0N/C( )8.99×109N⋅m2C2=56pC
22-2Chapter 22: Electric FieldsProblem 8(a) In Fig. 22-31, two point charges q1= -5q and q2= +2q are fixed to the x axis. (a) As amultiple of distance L, at what coordinate on the axis is the net electric field of the particleszero? (b) Sketch the electric field lines.Lxyq1q2Figure 22-31Answer(a) 1.72L to the right of q2.Solution(a)We can set up the equations for this problem if we will first determine possibleregions in which the electric field could be zero. Referring to Diagram 8a, note first of allthat the field cannot be zero accept along the line that runs through the two point charges.Furthermore by looking at the directions and relative sizes of the electric fields due to thetwo charges individually in three intervals along this line we see that any points where thefield is zero must lie to the right of q2.Lxyq1q2Eq1q2Eq2Eq2EEq1Eq1Diagram 8aTherefore if the field is zero it is at a point (x,0) where x > L. The total electric field at sucha point isEtot=14πεoq1r13r1+14πεoq2r23r2where r1is the vector from q1to the point (x,0) and r2is the vector from q2to the point(x,0). Which means that r1= xiand r2= (x - L)i. Therefore if the total electric field is zerothe previous equation for Etotbecomes0=14πεo−5qx3xi+14o2qx−L( )3x−L( )ior0=−5x2+2x−L( )2or3x2– 10Lx + 5L2= 0.
Chapter 22: Electric Fields22-3Solving this quadratic equation for x yields x = 0.61L or x = 2.72L. According to ourprevious analysis x must be greater than L therefore we find that the field will be zero at apoint 1.72L to the right of the q2.(b)A sketch of the field lines is shown in Diagram 8b. Notice in the sketch how thefield is more strongly affected by the -5q charge. For quite a distance from the -5q chargethe field still looks like that due to a single point charge as if the +2q charge were notpresent. In the vicinity of the +2q charge however the field deviates from the radial field
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