Question

Two charges are fixed on the x-axis. Charge q1 is +4.0 uC and is fixed at x = 0.0 cm. Charge 2 is fixed at x = 18 cm, but the value of its charge is unknown. Measurements indicate that the electric field is zero at x = 36 cm. Determine the magnitude and sign of charge q2.

Answer 1

To solve this problem, we can use the formula for the electric field created by a point charge:

E = k * |q| / r^2

where E is the electric field, k is the Coulomb constant (8.99 x 10^9 N m^2/C^2), |q| is the magnitude of the charge, and r is the distance between the charge and the point where the electric field is being measured.

At x = 18 cm, the electric field is given by:

E = k * |q1| / (18 cm)^2

At x = 36 cm, the electric field is given by:

E = k * |q1| / (36 cm)^2 + k * |q2| / (18 cm)^2

Since the total electric field at x = 36 cm is zero, we can set these two expressions equal to each other and solve for |q2|:

0 = k * 4.0 uC / (36 cm)^2 + k * |q2| / (18 cm)^2

0 = k * 4.0 uC / 1296 cm^2 + k * |q2| / 324 cm^2

0 = 8.99 x 10^9 N m^2/C^2 * 4.0 x 10^-6 C / 1296 x 10^-4 m^2 + 8.99 x 10^9 N m^2/C^2 * |q2| / 324 x 10^-4 m^2

0 = 0.028 N/C + 2.773 N/C * |q2|

-0.028 N/C = 2.773 N/C * |q2|

|q2| = 0.0101 C

Therefore, the magnitude of charge q2 is 10.1 uC. Since the electric field at x = 36 cm is zero, the charge q2 must be negative to cancel out the electric field created by charge q1. Thus, the charge q2 is -10.1 uC.