TP 20.30

 (a)    Given: q1 = -2.205 mC   @ (3.055 m, 4.501 m)

                    q2 = 1.800 mC   @ (-2.533 m, 0 m)

 

Find: V @ (0,0) - ? 

Let’s find the distance from the first charge to the origin:

r1 = (x2 + y2)1/2= ((3.055 m)2 + (4.501 m)2)1/2= 5.44 m

Electric potential due to charge 1 at the origin:

V1 = kq1/r1= (8.99 x 109 N-m2/C2)( -2.205 x 10-6 C)/(5.44 m) = - 3.64 x 103 V

Let’s find the distance from the second charge to the origin:

R2 = (x2 + y2)1/2= ((-2.533 m)2 + (0 m)2)1/2= 2.533 m

Electric potential due to charge 2 at the origin:

V2 = kq2/r2= (8.99 x 109 N-m2/C2)( 1.800 x 10-6 C)/(2.533 m) = 6.39 x 103 V

By superposition principle

V = V1 + V2 = - 3.64 x 103 V + 6.39 x 103 V = 2.75 x 103 V