Question

As clothing tumble in a dryer, they can become charged. If a small piece of lint with a charge of +1.62 E−19 C is attracted to the clothing by a force of 2.0 E−9 N, what is the magnitude of the electric field at this location?

0.38 E10 N/C
1.2 E10 N/C
3.2 E10 N/C
3.6 E10 N/C

Answer 1

1.2 E10 N/C

Step-by-step explanation:

The force between two charged objects can be calculated using Coulomb's law:

F = k * q1 * q2 / r^2

where F is the force, k is Coulomb's constant (k = 8.99 x 10^9 N m^2 / C^2), q1 and q2 are the charges of the two objects, and r is the distance between them.

Rearranging this equation to solve for the electric field at a point, we get:

E = F / q

where E is the electric field strength and q is the charge at that point.

Substituting the given values, we get:

E = (2.0 x 10^-9 N) / (1.62 x 10^-19 C)

E = 12345.68 N/C

Therefore, the magnitude of the electric field at this location is 1.235 x 10^4 N/C.