Physics 230 Spring 2002

Class Exercise Feb 1Question: A charge Q = 5m C is at x = 3m, y = 0. A second charge q = -3 mC is at x = 0, y = 4 m. Find the electric field at point P with coordinates x = 3 m, y = 4 m.
Solution1. Draw a clear diagram showing point P and the two fields (due to Q and q respectively) .


2. Find the electric field at P due to Q.
 E = kQ/r2 =(9x109N.m2/C2)(5x10-6C)/(4m)2=2.8x103N/C and the direction is the positive y-direction.

3. Find the electric field at P due to q.
E = kq/r2 =(9x109N.m2/C2)(3x10-6C)/(3m)2=3x103N/C and the direction is the negative x-direction.
 

4. Add the two contributions to find the total electric field at P.

The length of the vector is given by:
E2 = (2.8x103 N/C)2 + (3x103 N/C)2 = 16.84x106(N/C)2
Thus E = 4.1x103 N/C.
The direction is given by tan q =2.8/3 = 0.93 and thus q = 43 degrees.