Question

Two point charges are on the y axis. A 3.90-µC charge is located at y = 1.25 cm, and a -2.4-µC charge is located at y = −1.80 cm. (a) Find the total electric potential at the origin. V (b) Find the total electric potential at the point whose coordinates are (1.50 cm, 0). V

Answer 1

a) 1.6*10^6 V

b) 13.35*10^6 V

Step-by-step explanation:

The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:

`V=k(q_1)/(r_1)+k(q_2)/(r_2)` (1)

q1 = 3.90µC = 3.90*10^-6 C

q2 = -2.4µC = -2.4*10^-6 C

r1 = 1.25 cm = 0.0125 m

r2 = -1.80 cm = -0.018 m

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

You replace all the parameters in the equation (1):

`V=k[(q_1)/(r_1)+(q_2)/(r_2)]\\\\V=(8.98*10^9Nm^2/C^2)[(3.90*10^(-6)C)/(0.0125m)+(-2.4*10^(-6)C)/(0.018m)]=1.6*10^6V`

hence, the total electric potential is approximately 1.6*10^6 V

b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:

r1 = 0.0150m - 0.0125m = 0.0025m

r2= 0.015m + 0.018m = 0.033m

Then, you replace in the equation (1):

`V=(8.98*10^9Nm^2/C^2)[(3.90*10^(-6)C)/(0.0025m)+(-2.4*10^(-6)C)/(0.033m)]=13.35*10^6V`

hence, for y = 1.50cm you obtain V = 13.35*10^6 V