(Parte 11 de 11)

(a) Express the condition that must be satisfied if the net force on the egFF=

Chapter 21 46 proton is zero: Use Newton’s law of gravity and

Coulomb’s law to substitute for Fg and Fe:

Solve for m to obtain: G kem=

Substitute numerical values and evaluate m:

(b) Express the ratio of Fe and Fg:

Gm ke rke =

Substitute numerical values to obtain:

64 •• Picture the Problem The locations of the charges q1, q2 and q2 and the points at which we are calculate the field are shown in the diagram. From the diagram it is evident that Er along the axis has no y component. We can use Coulomb’s law for Er due to a point charge and the superposition principle to find Er at points P1 and P2. Examining the distribution of the charges we can see that there are two points where E = 0. One is between q2 and q3 and the other is to the left of q1.

The Electric Field 1: Discrete Charge Distributions 47

Using Coulomb’s law, express the electric field at P1 due to the three charges:

i E

P qqqP

rqrqr qk rkqrkqr kq

Substitute numerical values and evaluate 1PEr :

Express the electric field at P2: i qqqP

rqrqr qk

Substitute numerical values and evaluate 2PEr :

Letting x represent the x coordinate of a point where the magnitude of the electric field is zero, express the condition that E = 0 for the point between x = 0 and x = 1 cm:

P rqrqr qkE or

For x < −1 cm, let y = −x to obtain: ()()0 cm1

Chapter 21 48

Picture the Problem The locations of the charges q1, q2 and q2 and the point P2 at which we are calculate the field are shown in the diagram. The electric field on the x axis due to the dipole is given by 3dipole2xkpErr= whereip21aq=r . We can use Coulomb’s law for Er due to a point charge and the superposition principle to find Er

Express the electric field at P2 as the sum of the field due to the dipole and the point charge q2:

i E

qx aqx k xkqx aqk xkqx qP r where a = 1 cm.

Substitute numerical values and evaluate 2PEr :

64. Problem ofwith that agreement excellent in isresult this,than greatermuch not is i.e., interest, ofpoint the todistance theof 10% thanmore is dipole theof charges two theof separation theWhile a x

*6 •• Picture the Problem We can find the percentage of the free charge that would have to be removed by finding the ratio of the number of free electrons ne to be removed to give the penny a charge of 15 µC to the number of free electrons in the penny. Because we’re assuming the pennies to be point charges, we can use Coulomb’s law to find the force of repulsion between them.

(a) Express the fraction f of the free charge to be removed as the quotient of the number of electrons to be removed and the number of free

N nf e=

The Electric Field 1: Discrete Charge Distributions 49 electrons:

Relate N to Avogadro’s number, the mass of the copper penny, and the molecular mass of copper:

MmN N =A

M mNNA=

Relate ne to the free charge Q to be

e Qn−=e

A meN

M mN

Substitute numerical values and evaluate f:

(b) Use Coulomb’s law to express the force of repulsion between the two pennies:

2 r enkr

Substitute numerical values and evaluate F:

67 •• Picture the Problem Knowing the total charge of the two charges, we can use Coulomb’s law to find the two combinations of charge that will satisfy the condition that both are positive and hence repel each other. If just one charge is positive, then there is just one distribution of charge that will satisfy the conditions that the force is attractive and the sum of the two charges is 6 µC.

(a) Use Coulomb’s law to express the repulsive force each charge exerts on the other:

qkqF=

Express q2 in terms of the total

(Parte 11 de 11)

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