Hola

24-16. (Cont.) Ex = (2.70 x 105 N/C) cos 45.60 – (4.05 x 105 N/C) cos 45.60 = -9.45 x 104 N/C
Ey = (2.70 x 105 N/C) sin 45.60 – (4.05 x 105 N/C) sin 45.60 = +4.82 x 105 N/C
;ER = 4.91 x 105 N/C
;  = 78.90 N of W; ER = 4.91 x 105 N/C, 101.10

*24-17. A +4 nC charge is placed at x = 0 and a +6 nC charge is placed at x = 4 cm on an x-axis. Find thepoint where the resultant electric field intensity will be zero?


x = 1.80 cm

Applications of Gauss’s Law
24-18. Use Gauss’s law to show that thefield outside a solid charged sphere at a distance r from its center is given by

where Q is the total charge on the sphere.
Construct aspherical gaussian surface around the charged
sphere at the distance r from its center. Then, we have


24-19. Acharge of +5 nC is placed on the surface of a hollow metal sphere whose radius is 3 cm. Use Gauss’s law to find the electric field intensity at a distance of 1 cm from the surface of the sphere?What is the electric field at a point 1 cm inside the surface?
Draw gaussian surface of radius R = 3 cm + 1 cm = 4 cm.
This surface encloses a net positive charge of +5 nC and
has a surface areaof 4R2, so Gauss’ law gives us:
(a)
; E = 2.81 x 104 N/C, radially outward.

(b) Draw a gaussian surface just inside the sphere. Now, all charge resides on the surface of thesphere, so that zero net charge is enclosed, and oAE = q = 0.
E = 0, inside sphere

24-20. Two parallel plates, each 2 cm wide and 4 cm long, are stacked vertically so that the field intensitybetween the two plates is 10,000 N/C directed upward. What is the charge on each plate? First use Gauss’ law to find E between plates.
Draw gaussian cylinder of area A enclosing charge q....