Question

Two ping-pong balls, each with a positive charge of 2.5 x 10^-5 C, are located 9.0 cm apart. Determine the size of the electric force between them. Would this be enough force to support a 70 kg person against the downward force of gravity?

Answer 1

Answer on Question#65463, Physics / Astronomy | Astrophysics

693.7 N; enough

Question

Two ping-pong balls, each with a positive charge of $2.5 \times 10^{-5} C$, are located $9.0 \, \text{cm}$ apart. Determine the size of the electric force between them. Would this be enough force to support a $70 \, \text{kg}$ person against the downward force of gravity?

Solution

According to the Coulomb's law, electric force:

$$F_{C} = k \frac{q_{1} q_{2}}{r^{2}}$$

$k \approx 8.99 \times 10^{9} N \, m^{2} C^{-2}$ — Coulomb's constant, $q_{1} = q_{2} = 2.5 \times 10^{-5} C$, $r = 9.0 \, \text{cm} = 0.09 \, \text{m}$

Then,

$$F_{C} = 8.99 \times 10^{9} \frac{(2.5 \times 10^{-5})^{2}}{0.09^{2}} = \frac{8.99 \cdot 2.5^{2}}{9.0^{2}} \times \frac{10^{9} \cdot 10^{-10}}{10^{-4}} = \frac{8.99 \cdot 6.25}{81.0} \times 10^{3} \approx 693.7 \, N$$

Force of gravity can be expressed as:

$$F_{g} = m g$$

$m$ — mass, $g$ — acceleration of the free fall.

If we consider person standing somewhere on Earth, then $g \approx 9.81 \, m \, s^{-2}$

Hence,

$$F_{g} = 70 \cdot 9.81 = 686.7 \, N$$

$F_{C} > F_{g}$, thus, such size of the electric force is sufficient to support a $70 \, \text{kg}$ person against the downward force of gravity (on Earth).

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