Class Exercise Feb 1__Question__: 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.
__Solution__1. 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

3. Find the electric field at P due to *q.*
*E = kq/r ^{2 }=*(9x10

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

The length of the vector is given by:

E^{2 }= (2.8x10^{3} N/C)^{2 }+ (3x10^{3
}N/C)^{2 }= 16.84x10^{6}(N/C)^{2}

Thus E = 4.1x10^{3} N/C.

The direction is given by tan q =2.8/3 =
0.93 and thus q = 43 degrees.