Question

a 2.0-mm-diameter glass bead is positively charged. the potential difference between a point 2.0 mm from the bead and a point 4.0 mm from the bead is 500 v. what is the charge on the bead?

Answer 1

The charge on the bead is approximately 0.000222 µC.

Step-by-step explanation:

The potential difference between two points in an electric field is given by the equation:

V = k * q / r

where V is the potential difference, k is Coulomb's constant (9 x 10^9 N m^2/C^2), q is the charge, and r is the distance between the points.

In this case, the potential difference is 500 V, the distance from the bead is 2.0 mm (0.002 meters), and the distance from the bead is 4.0 mm (0.004 meters). We can rearrange the equation to solve for the charge:

q = V * r / k

Substituting in the values, we have:

q = (500 V) * (0.004 m) / (9 x 10^9 N m^2/C^2) = 0.000222 µC

Therefore, the charge on the bead is approximately 0.000222 µC.