Electrostatic Force. The magnitude of the electrostatic force between two point charges Q and q of the same sign is given by F(x)=kQq / x^2 where x is the distance (measured in meters) between the charges and k=9⋅10^9 Nm^2 / C^2 is a physical constant (C stands for coulomb, the unit of charge; N stands for newton, the unit of force).

1. Find the instantaneous rate of change of the force with respect to the distance between the charges.

2. For two identical charges with Q = q = 1C, what is the instantaneous rate of change of the force at a separation of x = 0.001m?

3. Does the magnitude of the instantaneous rate of change of the force increase or decrease with the separation? Explain

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debrubs97 debrubs97 · Answer 1 · 2019-10-15T17:38:50+02:00

Answer:

1-[tex]\frac{-2kQq}{x^{3} }[/tex]

2-18*10^18

3-Decrease

Step-by-step explanation:

1-The instantaneous rate of change of the force is the derivate

So: F'=kQq(-2x^-3)=[tex]\frac{-2kQq}{x^{3} }[/tex]

Q, q and k are constants, so we can apply the power rule to compute the derivate

2-By sustitution, we get

F'=[tex]\frac{-2*9*10^9*1*1}{0.001^{3} }[/tex]=18*10^18

3- The magnitude will decrease with the separation because we are dividing a constant by larger and larger values.