Question

centers of two small particles charged particles are separated by a distance of 1.2x10-4 meter. The charges on the particle are 8x10-19 coloumb, respectively. Calculate the magnitude of the electrostatic force between these two particles. Please help, I need to get a good grade for physics.

Answer 1

Data:
r (distance) = $1.2*10^{-4}m$
charges on the particle (Q*q) = $8*10^{-19}\:Coulomb$
k (Coulomb\:constant) = $8.988 * 10^9 N*m2/C2 \to K \approx 9.00 * 10^9 N*m2/C2$
F (magnitude) = ?

In applying the Coulomb Law we have:
$F = \frac{k*Q*q}{r^2}$

Solving:
$F = \frac{k*Q*q}{r^2}$
$F = \frac{9.00 * 10^9*8*10^{-19}}{(1.2*10^{-4})^2}$
$F = \frac{72 * 10^{9-19}}{(1.2*10^{-4})^2}$
$F = \frac{72 * 10^{-10}}{1.44*10^{-8}}$
$F = 50*10^{-10-(-8)}$
$F = 50*10^{-10+8}$
$F = 50*10^{-2}$
$\boxed{\boxed{F = 5.0*10^{-1}\:N}}}\end{array}}\qquad\quad\checkmark$

Note: Response used with approximate values