Chapter 22. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere.


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1 Chapte.3 What is the magnitude of a point chage whose electic field 5 cm away has the magnitude of.n/c. E E C.5 An atom of plutonium39 has a nuclea adius of 6.64 fm and atomic numbe Z94. Assuming that the positive chage is distibuted unifomly within the nucleus, what ae the magnitude and diection of the electic field at the suface of the nucleus due to the positive chage. Outside a unifomly chaged sphee, the field looks like that of a point chage at the cente of the sphee. E C ( m) N / C!.8 In Fig. 31, paticle 1 of chage 1 5 and paticle of chage + ae fixed to an xaxis (a) As a multiple of distance L, at what coodinate on the axis is the net electic field of the paticles zeo? (a) Sketch the net electic field lines. E1 E 1 (x,)
2 E 1 E 1 x E E 1 5 x (x L) 5 x (x L) 5 x (x L) x 1 (5 + 1)L 3.11 Figue 34 shows two chaged paticles on an x axis: C at x 3.m and C at x +3.m. What ae the (a) magnitude and (b) diection (elative to the positive diection of the x axis) of the net electic field poduced at point P at y +4.m x (., 4.) (3.,.) (+3.,.) We can do this poblem most easily by using the full vecto fom fo the electic field due to a point chage. ( 3.) i ˆ + (4. ) ˆ j m ˆ + + 3ˆ i + 4 ˆ j 5 ( (3))ˆ i + (4 ) ˆ j ˆ (3) + 4 5m 3ˆ i + 4 ˆ j 5
3 E E + + E C 4 π ε (5m) 3 ˆ 5 i + 4 ˆ j C 4 π ε (5m) 6 i ˆ + ˆ j N / C ˆ i C 4 π ε (5m) 3 ˆ i + 4 ˆ j 5.13 In Fig. 36, thee paticles ae fixed in place and have chages 1 +e and 3 +e. Distance a 6.µm. What ae the (a)magnitude and (b) diection of the net electic field at point P due to the paticles. 1 E E 3 E 1 3 The fields due to chages 1 and cancel exactly. This leaves us only to calculate the field due to 3. 1 a + a a E N / C at 45 a /.19. Find the magnitude and diection of the electic field at point P due to the electic dipole. You may assume that >>d
4 (,d/ ) + (, )  (,d/ ) We begin by witing the exact field. ( ) i ˆ + ( d ) ˆ j ( ) i ˆ + ( ( d )) ˆ j + ( d ) + ( d ) E E + + E ˆ + ˆ ˆ + + i ˆ d ˆ j + ( d ) ˆ i ˆ + d ˆ j + ( d ) We can substitute in. In the last step, since d/ is vey small we ignoe this tem. emembe that the diection of p is in the +j diection in this poblem (fom  to + chage).
5 E E + + E ˆ + ˆ ( + ( d ) ) i ˆ d ˆ j + ( d ) ( + ( d ) ) i ˆ + d ˆ j + ( d ) ( indicates downwad) d ( + ( d ˆ j ) ) 3/ p 3 (1 + ( d ˆ j ) ) 3/ p 3.4 In Fig 44, a thin glass od foms a semicicle of adius 5.cm. Chage is unifomly distibuted along the od, with pc in the uppe half and 4.5 pc in the l lowe half. What ae the (a) magnitude and (b) diection elative to the positive diection of the x axis of the electic field E a P, the cente of the cicle. Because of the symmety in the poblem, we can see that the net field will point downwad. We also can see that the contibution fom the bottom uate cicle is eual to the contibution fom the top uate cicle. Because of this, we only need to compute the downwad component due to the top uate cicle and multiply by. d d E We begin by defining a chage pe unit length π /
6 We now find the component of inteest and integate... d de cosθ de d d dθ The last facto of is because thee ae two uate ings. π / π / d dθ cosθ π / π 8ε tot 4ε cosθ dθ.5 In Fig 45, two cuved plastic ods, one of chage + and one of chage , fo a cicle of adius in the XY plane.. Te x axis passes though thei connection points and the chage is distibuted unifomly on both ods. What ae the magnitude and diection of the electic field E poduces at P, the cente of the cicle. Because of the symmety in the poblem, we can see that the net field will point downwad. We also can see that the contibution fom the bottom half cicle is eual to the contibution fom the top half cicle. Because of this, we only need to compute the downwad component due to the top half cicle and multiply by. d θ θ de We begin by defining a chage pe unit length
7 π We now find the component of inteest and integate... The last facto of is because thee ae two half ings. d de cosθ de d d dθ π / π / π / π / d dθ cosθ πε tot π / π / πε cosθ dθ.6
8 i de dez i z { Because of the symmety, we only need to compute the field along the axis (called z). de z de cosθ de d + z d 1 dϕ cosθ z + z E z E z π π π de z 1 dϕ ( + z ) 1 z dϕ ( + z ) 3/ 1 z ε ( + z ) 3/ z + z Now that we know the field, we take the deivative with espect to z and set it eual to zeo to find the location of the maximum field. m E z 1z f ( + z ) 3/ de z dz m 1(  z ) f ( + z ) 5/ z 1
9 3. In Fig. 51, positive chage 7.81pC is spead unifomly along a thin nonconducting od of length L 14.5cm. What ae the (a) magnitude and (b) diection (elative to the x axis of the electic field poduced at a distance 6.cm fom the od along its pependicula bisecto. (,d) L / L / Fist we wite d fo a little length of chage We then wite the,, ˆ fo the chage d. d dx ( x) + (d ) ( x) i ˆ + (d ) ˆ j ˆ x i ˆ + d ˆ j x + d We can then compute the x and y components of the Electic Field. You may need dx x (a + x ) 3/ a a + x
10 E E x d ˆ d dx d (x + d ) 3/ dx x i ˆ + d ˆ j (x + d ) x + d d L / d + L / 4 L / d + L / 4 d L d + L / 4 πε d L 4d + L dx d (x + d ) 3/ d x d + x L / L / Now we can compute with numbes C C / m.145m d.6m C / m πε.6m.145m 4(.6m) + (.145m) 1.43N / C.33 In Fig 5, a semi infinite non conducting od (that is, infinite in one diection only), has unifom linea chage density. Show that the electic field Ep at point P makes an angle of 45 degees with the od and this esult is independent of the the distance. x d L d E (, )
11 ( x) + ( ) ( x) i ˆ + ( ) ˆ j ˆ x i ˆ ˆ j x + We can then compute the x and y components of the Electic Field. You may need dx (a + x ) 3/ x a a + x We can compute each integal and then take the limit as L goes to infinity. We find that in this limit, the components ae identical fo all. This means that the atio of these components is 1 and the angle is 45 degees fo all. E E x d ˆ dx (x + ) lim :L E x 1 dx (x + ) 1 + L 1 dx (x + ) 3/ L + L L + L lim L x i ˆ ˆ j x + x x + x dx (x + ) 3/ dx (x + ) 3/ 1 + x x + x L L.34 This poblem is woked out in detail in section 37 of the book. We also woked this poblem in detail in class. Please eview the deivation thee.
12 σ C / m.5 1 m z 1 1 m E z σ ε (1 z z + ) N / C.39 An electon is eleased fom est in a unifom electic field of magnitude. # 1 4 N/C. Calculate the acceleation of the electon (ignoe gavitation). F ma E 1.6 # 1 a m E 19 C 9.1 # 131 $. # 14 N/C 3.5 # 1 kg 15 m/s.47 In Millikan s expeiment, an oil dop of adius 1.64 µm and density ρ.851g / cm 3 is suspended in chambe C (Fig. 14) when a downwad electic field E N / C is applied. Find the chage on the dop in tems of e F E If the doplet is suspended, we know that the net foce is zeo. This means that the weight is balanced by an upwad electic foce. We wite this balance mg F E mg E mg mg E We can wite the mass in tems of the volume of the dop.
13 m ρ 4 3 π g cm π ( cm) g kg mg E kg 9.8m / s C N / C 5e
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