A charge of #5 C# is at #(-6, 3 )# and a charge of #-2 C# is at #(-5 , 7 ) #. If both coordinates are in meters, what is the force between the charges?

Question

A charge of #5   C# is at #(-6, 3  )# and a charge of #-2  C# is at #(-5 ,  7  )  #. If both coordinates are in meters, what is the force between the charges?

Nathan Ashcroft · Accepted Answer

Recall that Coulomb's law tells you the electric force of attraction or repulsion between two point charges:

#bb(F_E = (kq_1q_2)/(r^2) = (q_1q_2)/(4piepsilon_0r^2))#

where #k = 1/(4piepsilon_0)# is a constant, #epsilon_0 = 8.854xx10^(-12) "C"^2"/N"cdot"m"^2# is the vacuum permittivity, and #r# is the distance between the two point charges.

Since you were given coordinates, recall the distance formula to calculate #r#:

#r = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

#= sqrt((-5 - -6)^2 + (7 - 3)^2)#

#= sqrt((1)^2 + (4)^2) = sqrt17#,

or, #color(green)(r^2 = 17)#.

The constant #k# is:

#color(green)(k) = 1/(4piepsilon_0) = 1/((4)(3.1415926535cdots)*(8.854xx10^(-12) "C"^2"/N"cdot"m"^2))#

#= color(green)(8.987xx10^9)# #color(green)("N"cdot"m"^2"/C"^2)#

So the electric force #F_E# is:

#color(blue)(F_E) = (kq_1q_2)/(r^2)#

#= ((8.987xx10^9"N"cdotcancel"m"^2"/"cancel("C"^2))(5 cancel"C")(-2 cancel"C"))/(17 cancel("m"^2))#

#= color(blue)(-5.287xx10^9)# #color(blue)("N")#

This negative value means that the two point charges are highly attracted to each other. This makes sense because they're oppositely-signed charges.

Benjamin Asiedu · Answer

undefined To find the force between the charges, you can use Coulomb's law, which states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Let's denote the positions of the charges as follows:
- Charge 1 with $ q_1 = 5 \, \text{C} $ is located at $ (-6, 3) $.
- Charge 2 with $ q_2 = -2 \, \text{C} $ is located at $ (-5, 7) $.

First, calculate the distance between the charges using the distance formula:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

$$ d = \sqrt{(-5 - (-6))^2 + (7 - 3)^2} $$

$$ d = \sqrt{(1)^2 + (4)^2} $$

$$ d = \sqrt{1 + 16} $$

$$ d = \sqrt{17} $$

Now, use Coulomb's law to find the force:

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

Where:
- $ k $ is Coulomb's constant, approximately $ 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 $.

Plugging in the given values:

$$ F = \frac{(8.99 \times 10^9) \cdot |5 \cdot (-2)|}{(\sqrt{17})^2} $$

$$ F = \frac{(8.99 \times 10^9) \cdot 10}{17} $$

$$ F = \frac{8.99 \times 10^{10}}{17} $$

$$ F \approx 5.29 \times 10^9 \, \text{N} $$

So, the force between the charges is approximately $ 5.29 \times 10^9 \, \text{N} $.