Question

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.

Answer 1

(a) 1745.36 V

(b) 8312.5 N/C

Step-by-step 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

`V=(Kq)/(r)`

The electric potential at A due to charge q1 is

`V_(1)=(Kq_(1))/(r_(1))`

`V_(1)=(9* 10^(9)* 8.6* 10^(-9))/(0.14)`

V1 = 552.86 V

the electric potential at A due to charge q2 is

`V_(2)=(Kq_(2))/(r_(2))`

`V_(2)=(9* 10^(9)* 5.3* 10^(-9))/(0.04)`

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

`E = (Kq)/(r^(2))`

The electric field due to charge q1 at B is

`E_(1) = (Kq_(1))/(r_(1)^(2))` towards right

`E_(1) = (9* 10^(9)* 8.6* 10^(-9))/(0.06^(2))`

E1 = 21500 N/C towards right

The electric filed due to charge q2 at B is

`E_(2) = (Kq_(2))/(r_(2)^(2))` towards left

`E_(2) = (9* 10^(9)* 5.3 * 10^(-9))/(0.04^(2))`

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