A point charge q1=8.60 nC is fixed at the origin. A second point charge q2=5.30 nC is fixed at the point x=10.0 cm. (a) Find the electric potential at point A, located at xA=14.0 cm. (b) Find the total electric field (magnitude and direction) at point B, located at xB=6.00 cm.

(a) 1745.36 V

(b) 8312.5 N/C

Explanation:

q1 = 8.6 nC at origin

q2 = 5.3 nC at x = 10 cm

(a)

Let the electric potential at A is V.

The formula for the electric potential is given by

The electric potential at A due to charge q1 is

V1 = 552.86 V

the electric potential at A due to charge q2 is

V2 = 1192.5 V

The total potential at A is

V = V1 + V2 = 552.86 + 1192.5 = 1745.36 V

(b)

The formula for the electric field is given by

The electric field due to charge q1 at B is

  towards right

E1 = 21500 N/C towards right

The electric filed due to charge q2 at B is

  towards left

E2 = 29812.5 N/C towards left

ne electric filed at B is

E = E2 - E1

E = 29812.5 - 21500 = 8312.5 N/C towards origin

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.