Question

Two identical spheres, A and B, are placed 15 cm apart. The sphere A is charged to 5.0 μC and the sphere B is charged to –7.0 μC.
a. How large is the force between the two spheres?
b. Is the force repulsive or attractive?
ASAPPPPPP

Answer 1

Answer:

a. F = 14 N

b. Since the spheres' charges have different signs, the force is attractive

Explanation:

Coulomb's Law

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\cdot\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

The spheres A and B have charges of:

$q_1=5\cdot 10^{-6}\ c$

$q_2=7\cdot 10^{-6}\ c$

The distance is d=15 cm = 0.15 m.

a. The force is now calculated:

$\displaystyle F=9\cdot 10^9\cdot\frac{5\cdot 10^{-6}\cdot 7\cdot 10^{-6}}{0.15^2}$

$\displaystyle F=9\cdot 10^9\cdot\frac{35\cdot 10^{-12}}{0.0225}$

F = 14 N

b. The electrostatic force between charges is repulsive if they have like signs, the force between them is attractive if they have opposite signs.

Since the spheres' charges have different signs, the force is attractive