A charge of #-2 C# is at #(6 , -5 )# and a charge of #-4 C# is at #(-3 , -9 ) #. If both coordinates are in meters, what is the force between the charges?

Answer 1

The force between the charges is #0.74xx10^9# Newtons.

According to the Coulomb's Law, the magnitude of the electric force exerted by a charge '#q_1#' on another charge '#q_2#' a distance '#r#' away is thus, given by #F_e=(k|q_1q_2|)/r^2# Here the value of '#k#' in SI units is #k= 8.987551787xx10^9 (N-m^2)/C^2# #k~~8.988xx10^9(N-m^2)/C^2# Using the given coordinates we can find of the distance between them: Distance#r=sqrt[(x_2-x_1)^2+(y_2-y_1)^2]# Calculating with given values we get , #r=sqrt(97)=9.84 m#
Substituting all given values in above force equation we get, #F_e={8.988xx10^9(N-m^2)/C^2}{[(-2C)(-4C)]/[sqrt(97)m]^2}# #:. F= 0.74xx10^9 # Newtons
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Answer 2

Use Coulomb's Law: ( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} ) where ( k ) is Coulomb's constant (8.99 × 10^9 N m²/C²), ( q_1 ) and ( q_2 ) are the charges, and ( r ) is the distance between them.

[ F = \frac{(8.99 \times 10^9) \cdot |-2 \cdot (-4)|}{(6 - (-3))^2 + ((-5) - (-9))^2} ]

Calculate to find the force.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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