1

Example continued

Field due to q1

E = 1010 N.m2/C2 10 X10-9 C/(3m)2 = 11 N/Cin the y direction.

Recall E =kq/r2

and k=8.99 x 109 N.m2/C2

x

y

q1=10 nc q2 =15 nc4

3

Ey= 11 N/C

Ex= 0

Field due to q2

5

E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C

at some angle φResolve into x and y components

φ

E

Ey=E sin φ = 6 ∗ 3/5 =18/5 = 3.6 Ν/C

Ex=E cos φ = 6 ∗ (−4)/5 =−24/5 = −4.8 Ν/C

φ

Now add all components

Ey= 11 + 3.6 = 14.6 N/C

Ex= -4.8 N/C

€

E = Ex2 + Ey

2

Magnitude

2

Example continued

xq1=10 nc q2 =15 nc

4

3

Ey= 11 + 3.6 = 14.6 N/C

Ex= -4.8 N/C

E

€

E = 14.6( )2

+ - 4.8( )2

=15.4N /C

Magnitude of electric field

€

E = Ex2 + Ey

2

φ1

φ1 = tan-1 Ey/Ex= atan (14.6/-4.8)= -72.8 deg

3

Motion of point charges in electric fields

• When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newton’s Law:

a = F/m = qE/m for uniform electric fields

a = F/m = mg/m = g for uniform gravitational fields

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.

Replace g by qE/m in all equations;

For example, In y =1/2at2 we get y =1/2(qE/m)t2