Question

The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far apart.

Answer 1

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

$F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}$

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ $\frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2} }{Distance_{old}^{2} }$

Now, we know the new distance is half the original distance,

$F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}$

$F_{new} = 4F_{old}$

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

Learn more about electrical force here

brainly.com/question/2526815

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