Question

Find the force between charges of +10.0 C and -50.0 C located 20.0 cm apart

Answer 1

The magnitude of the electrostatic force between the two charges is $1.12\cdot 10^{14} N$

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

$q_1 = +10.0 C$

$q_2 = -50.0 C$

r = 20.0 cm = 0.20 m

Therefore, the force between the charges is

$F=(8.99\cdot 10^9) \frac{(10)(-50)}{(0.20)^2}=-1.12\cdot 10^{14} N$

where the negative sign means that the force is attractive, since the two charges have opposite sign.

Learn more about electrostatic force here:

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