Respuesta :
To solve this problem we will use the Coulomb laws that define electrostatic force. Through this mathematical function we will derive the force as a function of time, which will give us in consideration of equality the one derived from the load with respect to time. The relationship for electrostatic force is given by the relationship,
[tex]F =\frac{kq_1q_2}{d^2}[/tex]
[tex]F=\frac{k*q^2}{d^2}[/tex]
Here,
k = Coulomb's Constant
q = Charge of the object (as they are the same, we have the value squared)
d = Distance
From the statement we have that the charge respect the time is,
[tex]\frac{dq}{dt} = 4.0*10^{-9} C/s[/tex]
And the derivative of the force respect to time is,
[tex]\frac{dF}{dt} = \frac{2kq(dq/dt)}{d^2}[/tex]
After 1 sec the charge on the particle is [tex]q = 4.0*10^{-9} s[/tex]
Replacing our values we have that,
[tex]\frac{dF}{dt} = 2\frac{(9*10^9)(4.0*10^{-9})^2}{(0.025^2)}[/tex]
[tex]\frac{dF}{dt} = 0.000468N/s[/tex]
Therefore the rate that the force between them increasing 1.0s after charging begins is [tex]4.68*10^{-4}N/s[/tex]
The force between two charges is increasing with rate of [tex]4.608*10^{-4} N/s[/tex]
The force between two point charges is given by Coulombs law.
[tex]F=k\frac{q_{1}q_{2}}{r^{2} } [/tex]
Where [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are two point charges and r is distance between them.
k is coulomb constant. [tex]k=9*10^{9}Nm^{2}/C ^{2} [/tex].
Given that, [tex]\frac{dq}{dt}=4nC/s [/tex]
Charge at t = 1 s , [tex]q=4nC=4*10^{-9}C [/tex]
Distance [tex]r=2.5cm=2.5*10^{-2}m [/tex]
Differentiate force expression with respect to time.
[tex]\frac{dF}{dt}=k\frac{2q*\frac{dq}{dt} }{r^{2} } [/tex]
Substitute values in above expression.
[tex]\frac{dF}{dt} =9*10^{9}*\frac{2*4*10^{-9} *4*10^{-9} }{(0.025)^{2} } \\ \\ \frac{dF}{dt}=4.608*10^{-4} N/s[/tex]
Hence., the force between two charges is increasing with rate of [tex]4.608*10^{-4} N/s[/tex]
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