A charge of #-2 C# is at #(1,4 )# and a charge of #-1 C# is at #(12 ,-5) #. If both coordinates are in meters, what is the force between the charges?

Question

A charge of #-2  C# is at #(1,4 )# and a charge of #-1  C# is at #(12  ,-5)  #. If both coordinates are in meters, what is the force between the charges?

Emily Abbate · Accepted Answer

#8.9*10^7N#

Distance between charges:

#|vec(r)| = sqrt(Deltax^2+Deltay^2) = sqrt(11^2+9^2) = sqrt(202)#

Using Coulomb's law:

#vec(F) = (Q_1Q_2)/(4piepsilon_0r^2)hat(r)#

The direction of the force is along the line connecting the charges radially and we will have #F_(12) = -F_(21)# anyway so the direction is not critical. We will find the magnitude now:

#|vec(F)| = 1/(4piepsilon_0)((-2)(-1))/(sqrt(202))^2 = 1/(2piepsilon_0(202)) = 8.9*10^7N#

Amelia Adams · Answer

undefined To find the force between the charges, we can use Coulomb's law:

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

Where:
- $ F $ is the force between the charges.
- $ k $ is Coulomb's constant, approximately $ 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 $.
- $ q_1 $ and $ q_2 $ are the magnitudes of the charges.
- $ r $ is the distance between the charges.

Given:
$ q_1 = -2 \, \text{C} $
$ q_2 = -1 \, \text{C} $
$ r = \sqrt{(12-1)^2 + (-5-4)^2} $

Calculate $ r $:
$ r = \sqrt{(11)^2 + (-9)^2} $
$ r = \sqrt{121 + 81} $
$ r = \sqrt{202} $

Calculate the force:
$ F = \frac{8.99 \times 10^9 \cdot |-2 \cdot -1|}{\sqrt{202}^2} $
$ F = \frac{8.99 \times 10^9 \cdot 2}{202} $
$ F = \frac{17.98 \times 10^9}{202} $
$ F = \frac{17.98}{202} \times 10^9 $
$ F ≈ 8.9 \times 10^7 \, \text{N} $

The force between the charges is approximately $ 8.9 \times 10^7 \, \text{N} $.