What is the force, in terms of Coulomb's constant, between two electrical charges of #24 C# and #32 C# that are #12 m # apart?

Question

Charlotte Aaron · Accepted Answer

If #q_1# and #q_2# are charges and #r# is the distance between them then the force #F# between them is given by #F=(kq_1q_2)/r^2# Where #k# is a constant and is known as Coulomb's force constant Here Let #q_1=24C#, #q_2=32C# and #r=12m#, #F=(k*24*32)/12^2# #implies F=(768k)/144=# #implies F=5.33kN# The force between the charges is repulsive because the force is positive.

Julia Aguilar · Answer

undefined The force between two charges can be calculated using Coulomb's law, which states:

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

where $ F $ is the force, $ k $ is Coulomb's constant ($ 8.9875 \times 10^9 $ N m²/C²), $ q_1 $ and $ q_2 $ are the magnitudes of the charges, and $ r $ is the distance between the charges.

Given that $ q_1 = 24 $ C, $ q_2 = 32 $ C, and $ r = 12 $ m, the force $ F $ is:

$$ F = (8.9875 \times 10^9 \text{ N m²/C²}) \cdot \frac{(24 \text{ C}) \cdot (32 \text{ C})}{(12 \text{ m})^2} $$

$$ F = (8.9875 \times 10^9 \text{ N m²/C²}) \cdot \frac{768 \text{ C²}}{144 \text{ m²}} $$

$$ F = (8.9875 \times 10^9 \text{ N m²/C²}) \cdot 5.3333 \text{ C²/m²} $$

$$ F = 4.8 \times 10^{10} \text{ N} $$

Therefore, the force between the two charges is $ 4.8 \times 10^{10} $ N.