Chapter 21The Electric Field 1: Discrete Charge DistributionsConceptual Problems.1. Similarities:Differences:The force between charges andmasses varies as 1/ r2.There are positive and negative charges butonly positive masses.The force is directly proportional tothe product of the charges ormasses.Like charges repel; like masses attract.The gravitational constant G is many ordersof magnitude smaller than the Coulombconstant k.2. Determine the Concept No. In order to charge a body by induction, it must have chargesthat are free to move about on the body.
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An insulator does not have such charges.3.Determine the Concept During this sequence of events, negative charges are attractedfrom ground to the rectangular metal plate B. When S is opened, these charges are trappedon B and remain there when the charged body is removed. Hence B is negatively chargedand correct. Is )( c4.( a) Connect the metal sphere to ground; bring the insulating rod near the metal sphereand disconnect the sphere from ground; then remove the insulating rod. The sphere willbe negatively charged.( b) Bring the insulating rod in contact with the metal sphere; some of the positive chargeon the rod will be transferred to the metal sphere.( c) Yes. First charge one metal sphere negatively by induction as in ( a).
Then use thatnegatively charged sphere to charge the second metal sphere positively by induction.1. Chapter 212.5.
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Determine the Concept Because the spheres are conductors, there are free electrons onthem that will reposition themselves when the positively charged rod is brought nearby.( a) On the sphere near the positivelycharged rod, the induced charge is negativeand near the rod. On the other sphere, thenet charge is positive and on the side farfrom the rod. This is shown in the diagram.( b) When the spheres are separated and farapart and the rod has been removed, theinduced charges are distributed uniformlyover each sphere. The charge distributionsare shown in the diagram.6. Determine the Concept The forces actingon + q are shown in the diagram. The forceacting on + q due to − Q is along the linejoining them and directed toward − Q. Theforce acting on + q due to + Q is along theline joining them and directed away from+ Q.Because charges + Q and − Q are equal in magnitude, the forces due to these charges areequal and their sum (the net force on + q) will be to the right and so correct.
Is )( e Notethat the vertical components of these forces add up to zero.7. Determine the Concept The acceleration of the positive charge is given by.0 EFarrrmqm Because q0 and m are both positive, the acceleration is in the samedirection as the electric field.
Is )( d.8. Determine the Concept Eris zero wherever the net force acting on a test charge iszero. At the center of the square the two positive charges alone would produce a netelectric field of zero, and the two negative charges alone would also produce a netelectric field of zero. Thus, the net force acting on a test charge at the midpoint of the. The Electric Field 1: Discrete Charge Distributions3square will be zero. Is )( b9.( a) The zero net force acting on Q could be the consequence of equal collinear chargesbeing equidistant from and on opposite sides of Q.( b) The charges described in ( a) could be either positive or negative and the net force onQ would still be zero.( c) Suppose Q is positive.
Imagine a negative charge situated to its right and a largerpositive charge on the same line and the right of the negative charge. Such an arrangementof charges, with the distances properly chosen, would result in a net force of zero actingon Q.( d) Because none of the above are correct, correct. Is )( d10. Determine the Concept We can use therules for drawing electric field lines todraw the electric field lines for this system.In the sketch to the right we’ve assigned 2field lines to each charge q.11.Determine the Concept We can use therules for drawing electric field lines todraw the electric field lines for this system.In the field-line sketch to the right we’veassigned 2 field lines to each charge q. Chapter 214.12.Determine the Concept We can use therules for drawing electric field lines todraw the electric field lines for this system.In the field-line sketch to the right we’veassigned 7 field lines to each charge q.13. Determine the Concept A positive charge will induce a charge of the opposite sign onthe near surface of the nearby neutral conductor. The positive charge and the inducedcharge on the neutral conductor, being of opposite sign, will always attract one another.correct.
Is )( a.14. Determine the Concept Electric field lines around an electric dipole originate at thepositive charge and terminate at the negative charge. Only the lines shown in ( d) satisfythis requirement. Is )( d.15. Determine the Concept Because θ ≠ 0, a dipole in a uniform electric field willexperience a restoring torque whose magnitude is θsin xpE. Hence it will oscillateabout its equilibrium orientation, θ = 0.
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The Electric Field 1: Discrete Charge Distributions5( e) True 17. Determine the Concept The diagramshows the metal balls before they areplaced in the water. In this situation, the netelectric field at the location of the sphereon the left is due only to the charge – q onthe sphere on the right. If the metal ballsare placed in water, the water moleculesaround each ball tend to align themselveswith the electric field.
This is shown forthe ball on the right with charge – q.( a) The net electric fieldrE net that produces a force on the ball on the left is thefieldrE due to the charge – q on the ball on the right plus the field due to the layerof positive charge that surrounds the ball on the right. This layer of positivecharge is due to the aligning of the water molecules in the electric field, and theamount of positive charge in the layer surrounding the ball on the left will be lessthan +q. Chapter 216.18. Determine the Concept Yes. A positively charged ball will induce a dipole on the metalball, and if the two are in close proximity, the net force can be attractive.19. Determine the Concept Assume that the wand has a negative charge.
When the chargedwand is brought near the tinfoil, the side nearer the wand becomes positively charged byinduction, and so it swings toward the wand. When it touches the wand, some of thenegative charge is transferred to the foil, which, as a result, acquires a net negative chargeand is now repelled by the wand. Estimation and Approximation20. Picture the Problem Because it is both very small and repulsive, we can ignore thegravitational force between the spheres.
It is also true that we are given no informationabout the masses of these spheres. The Electric Field 1: Discrete Charge Distributions9Substitute for n to obtain:qkTPWE σ= (1)Substitute numerical values and evaluate E:( )( )( )( )( )( )( ) N/C1041.2K300J/K1038.1C106.1J/eV106.1eV1N/m91925219×=×××= −−−−E( b) From equation (1) we see that: PE ∝ and 1−∝ TEi.e., E increases linearly with pressure andvaries inversely with temperature.23. Picture the Problem We can use Coulomb’s law to express the charge on the rod interms of the force exerted on it by the soda can and its distance from the can. We canapply Newton’s 2nd law in rotational form to the can to relate its acceleration to theelectric force exerted on it by the rod. The Electric Field 1: Discrete Charge Distributions11.27. Picture the Problem We can find the number of coulombs of positive charge there are in1 kg of carbon from, where nenQ C6= C is the number of atoms in 1 kg of carbon andthe factor of 6 is present to account for the presence of 6 protons in each atom.
The Electric Field 1: Discrete Charge Distributions1330. Picture the Problem The configuration ofthe charges and the forces on the fourthcharge are shown in the figure as is acoordinate system. From the figure it isevident that the net force on q4 is along thediagonal of the square and directed awayfrom q3. Chapter 2114Convert to: 4,3 rr4,3̂r( ) ( )( ) ( )jijirrrˆ707.0ˆ707.0m05.0m05.0ˆm05.0ˆm05.0ˆ224,34,34,3+= rrSubstitute numerical values and evaluate 4,3 Fr:( )( ) ( ) ( )( ) ( ) jijiFˆN1014.1ˆN1014.1ˆ707.0ˆ707.0m205.0nC3nC3/CmN104,3−− ×−×−=+⎟⎟⎠⎞⎜⎜⎝⎛−⋅×=rSubstitute and simplify to find 4 Fr:( ) ( ) ( ) ( )( ) ( ) jijiijFˆN1010.2ˆN1010.2ˆN1014.1ˆN1014.1ˆN1024.3ˆN104−−−−−−×+×=×−×−×+×=r31. Picture the Problem The configuration of the charges and the forces on q3 are shown inthe figure as is a coordinate system. From the geometry of the charge distribution it isevident that the net force on the 2 µC charge is in the negative y direction.
We can applyCoulomb’s law to express 3,1 Frand 3,2 Frand then add them to find the net force on q3.The net force acting on q3 is given by: 3,23,13 FFFrrr+=. The Electric Field 1: Discrete Charge Distributions15Express the force that q1 exerts onq3:jiF ˆsinˆcos3,1 θθ FF −=rwhere( )( )(( ) ( ))N3.12m0.08m0.03C2C5C/mN1031=+⋅×µµrqkqFand°=⎟⎟⎠⎞⎜⎜⎝⎛= − 6.20cm8cm3tan 1θExpress the force that q2 exerts onq3:jiF ˆsinˆcos3,2 θθ FF −−=rSubstitute for 3,1 Frand 3,2 Frandsimplify to obtain:jjijiFˆsin2ˆsinˆcosˆsinˆcos3θθθθθFFFFF−=−−−=rSubstitute numerical values andevaluate 3 Fr:( )( ) jjFˆN66.8ˆ6.20sinN3.1223−=°−=r.32. Picture the Problem The positions of thecharges are shown in the diagram. It isapparent that the electron must be locatedalong the line joining the two charges.Moreover, because it is negatively charged,it must be closer to the −2.5 µC than to the6.0 µC charge, as is indicated in the figure.We can find the x and y coordinates of theelectron’s position by equating the twoelectrostatic forces acting on it and solvingfor its distance from the origin.We can use similar triangles to express thisradial distance in terms of the x and ycoordinates of the electron.Express the condition that must be ee FF,2,1 =. Chapter 2116satisfied if the electron is to be inequilibrium:Express the magnitude of the forcethat q1 exerts on the electron:( )21,1m25.1+=rekqF eExpress the magnitude of the forcethat q2 exerts on the electron:22,2 reqkF e =Substitute and simplify to obtain:( ) 2221m25.1 rqrq=+Substitute for q1 and q2 andsimplify:( ) ( )0m25.1m2361.2m4.1 122=− −− rrSolve for r to obtain:m0.4386andm036.2−rrBecause r.
The Electric Field 1: Discrete Charge Distributions17.33. Picture the Problem Let q1 represent thecharge at the origin, q2 the charge at (0, 0.1m), and q3 the charge at(0.2 m, 0). The diagram shows the forcesacting on each of the charges. Note theaction-and-reaction pairs.
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