Question

What would be the electric field (magnitude and direction) of 1.50 cm to the right of a charge of -6.5 × 10-6 C?

Answer 1

Step-by-step explanation:

The formula for the electric field is

`E=(kQ)/(r^2)` where E is the magnitude of the electric field, k is Coulomb's constant, Q is the charge of the particle (which is NOT included in the formula), and r is the distance between the centers of the charges (for lack of a better description).

`E=((9.0*10^9)*(6.5*10^(-6)))/((1.50)^2)` and we get, to 2 sig fig's

E = 2.6 × 10⁴ to the left (since electric fields are always pointing toward the negative charge and the electric field is to the right of the negative charge)