Successfully reported this slideshow.
Upcoming SlideShare
×

# Electric field solving

1,589 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

### Electric field solving

1. 1. Example continued Field due to q 1 E = 10 10 N.m 2 /C 2 10 X10 -9 C/(3m) 2 = 11 N/C in the y direction. Recall E =kq/r 2 and k=8.99 x 10 9 N.m 2 /C 2 E y = 11 N/C E x = 0 Field due to q 2 5 E = 10 10 N.m 2 /C 2 15 X10 -9 C/(5m) 2 = 6 N/C at some angle  Resolve into x and y components  E E y =E sin  C E x =E cos  C  Now add all components E y = 11 + 3.6 = 14.6 N/C E x = -4.8 N/C Magnitude x y q 1 =10 nc q 2 =15 nc 4 3
2. 2. Example continued E y = 11 + 3.6 = 14.6 N/C E x = -4.8 N/C E Magnitude of electric field      tan -1 E y /E x = atan (14.6/-4.8)= -72.8 deg x q 1 =10 nc q 2 =15 nc 4 3
3. 3. Motion of point charges in electric fields <ul><li>When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newton’s Law: </li></ul><ul><li>a = F/m = qE/m for uniform electric fields </li></ul><ul><li>a = F/m = mg/m = g for uniform gravitational fields </li></ul><ul><li>If the field is uniform, we now have a projectile motion problem- constant acceleration in one direction. So we have parabolic motion just as in hitting a baseball, etc except the magnitudes of velocities and accelerations are different. </li></ul><ul><li>Replace g by qE/m in all equations; </li></ul><ul><li>For example, In y =1/2at 2 we get y =1/2(qE/m)t 2 </li></ul>