A charge of #-2 C# is at #(-5 , 1 )# and a charge of #-3 C# is at #(0 ,4) #. If both coordinates are in meters, what is the force between the charges?

Question

A charge of #-2 C# is at #(-5  , 1   )# and a charge of #-3 C# is at #(0  ,4)  #. If both coordinates are in meters, what is the force between the charges?

Nora Arzate · Accepted Answer

Magnitude of #vecF# : #4.46 xx 10^10 N#

Direction of #vecF_(1"on"2)# : #31.0^o#

Direction of #vecF_(2"on"1)# : #211.0^o# Coulomb's law, which describes the relationship between the strength of the electric force between two point charges and can be expressed as an equation, can be used to solve this problem. #F = 1/(4piepsilon_0)(|q_1q_2|)/(r^2)# The quantity #1/(4piepsilon_0)# is a proportionality constant, and can sometimes be simply called #k#, and has a value of #1/(4pi(8.854 xx 10^-12(C^2)/(N*m^2)))# # = 8.988 xx 10^9(N*m^2)/(C^2)# The variables #q_1# and #q_2# are, in no particular order, the magnitude of the electric charges of the two point charges, which are #-2C# and #-3C#. The variable #r# is the distance between the two point charges, which is #|sqrt((-5m)^2 + (1m)^2) - sqrt((0m)^2 + (4m)^2)| = 1.10m# The magnitude of the electric force between the two points is now known since we have all the necessary variables. #F = 8.988 xx 10^9(N*m^2)/(C^2)((|(-2C)(-3C)|)/((1.10m)^2))# #= color(blue)(4.46 xx 10^10 N# The force is repulsive because both point charges are negative, and it represents the amount of force that the particles exert against one another. If you want the direction of the force, which completes the vectoral description of the force of repulsion, we can use the fact that the force of object #1# acting on #2# is equal to the negative of the force of object #2# acting on #1#. If we let the point with charge #-2C# with coordinates #(-5,1)# be #color(orange)1#, and the point with charge #-3C# with coordinates #(0,4)# be #color(red)2# (for convenience), the direction of #vec F_(color(orange)1"on"color(red)2)# is #arctan((y_2-y_1)/(x_2-x_1)) = arctan((4m-1m)/(0m--5m)) = color(green)(31.0 ^o# The direction of #vec F_(2"on"1)# is simply the direction opposite to this in a circle: #180.0^o + color(green)(31.0^o) = color(purple)(211.0^o#

Adrian Ayala · Answer

undefined The force between the charges can be calculated using 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 m}^2/\text{C}^2 $),
- $ q_1 $ and $ q_2 $ are the magnitudes of the charges,
- $ r $ is the distance between the charges.

Given that $ q_1 = -2 \, \text{C} $, $ q_2 = -3 \, \text{C} $, and the coordinates are in meters, the distance $ r $ between the charges can be found using the distance formula:

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

Substituting the given coordinates, $ x_1 = -5 \, \text{m} $, $ y_1 = 1 \, \text{m} $, $ x_2 = 0 \, \text{m} $, $ y_2 = 4 \, \text{m} $, into the distance formula yields $ r = \sqrt{{5^2 + 3^2}} = \sqrt{34} \, \text{m} $.

Plugging the values into Coulomb's law:

$$ F = \frac{{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot |-2 \, \text{C} \cdot (-3 \, \text{C})|}}{{(\sqrt{34})^2}} $$

$$ F \approx \frac{{8.99 \times 10^9 \cdot 6}}{{34}} $$

$$ F \approx \frac{{53.94 \times 10^9}}{{34}} $$

$$ F \approx 1.587 \times 10^9 \, \text{N} $$

Therefore, the force between the charges is approximately $ 1.587 \times 10^9 \, \text{N} $.