{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

626_PartUniversity Physics Solution

# 626_PartUniversity Physics Solution - Electric Charge and...

This preview shows pages 1–3. Sign up to view the full content.

Electric Charge and Electric Field 21-19 The components of the two fields are shown in Figure 21.49g. 1 2 2 0 1 4 q E E r π = = P 9 9 2 2 1 2 6.00 10 C (8.988 10 N m /C ) (0.250 m) E × = × 1 2 862.8 N/C E E = = Figure 21.49g 1 1 2 2 cos , cos x x E E E E θ θ = − = + 1 2 0 x x x E E E = + = 1 1 2 2 sin , sin y y E E E E θ θ = + = + ( )( ) 1 2 1 1 2 2 sin 2 862.8 N/C 0.800 1380 N/C y y y y E E E E E θ = + = = = = 1380 N/C, in the -direction. E y = + E VALUATE : Point a is symmetrically placed between identical charges, so symmetry tells us the electric field must be zero. Point b is to the right of both charges and both electric fields are in the + x -direction and the resultant field is in this direction. At point c both fields have a downward component and the field of 2 q has a component to the right, so the net E G is in the 4th quadrant. At point d both fields have an upward component but by symmetry they have equal and opposite x -components so the net field is in the + y -direction. We can use this sort of reasoning to deduce the general direction of the net field before doing any calculations. 21.50. I DENTIFY : Apply Eq.(21.7) to calculate the field due to each charge and then calculate the vector sum of those fields. S ET U P : The fields due to 1 q and to 2 q are sketched in Figure 21.50. E XECUTE : 9 2 2 0 1 (6.00 10 C) ˆ ˆ ( ) 150 N/C 4 (0.6 m) π × = = − E i i G P . 9 1 2 2 0 1 1 1 ˆ ˆ ˆ ˆ (4.00 10 C) (0.600) (0.800) (21.6 28.8 )N C 4 (1.00 m) (1.00 m) π = × + = E i j i + j G P . 1 2 ˆ ˆ ( 128.4 N/C) (28.8 N/C) = − E = E + E i + j G G G . 2 2 (128.4 N/C) (28.8 N/C) 131.6 N/C E = + = at 1 28.8 tan 12.6 128.4 θ = = ° above the x axis and therefore 196.2 ° counterclockwise from the + x axis. E VALUATE : 1 E G is directed toward 1 q because 1 q is negative and 2 E G is directed away from 2 q because 2 q is positive. Figure 21.50 21.51. I DENTIFY : The resultant electric field is the vector sum of the field 1 1 2 2 of and of . q q E E G G S ET U P : The placement of the charges is shown in Figure 21.51a. Figure 21.51a

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
21-20 Chapter 21 E XECUTE : (a) The directions of the two fields are shown in Figure 21.51b. 1 1 2 2 0 1 1 4 q E E r π = = P 9 9 2 2 1 2 6.00 10 C (8.988 10 N m /C ) (0.150 m) E × = × 1 2 2397 N/C E E = = Figure 21.51b 1 1 2x 2 2397 N/C, 0 2397 N/C, 0 x y y E E E E = − = = − = 1 2 2( 2397 N/C) 4790 N/C x x x E E E = + = = − 1 2 0 y y y E E E = + = The resultant electric field at point a in the sketch has magnitude 4790 N/C and is in the -direction. x (b) The directions of the two fields are shown in Figure 21.51c.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern