What would the slope of the force between 2 charged particles vs 1/r^2 give me?

AI Thread Summary
The discussion revolves around understanding the slope of a graph depicting the force between two charged particles as a function of distance, specifically in relation to the inverse square law (1/r^2). Participants seek clarification on what to plot on the y-axis and x-axis, along with their respective units. The gradient of the plot, defined as Δy/Δx, raises questions about how to determine the units for the slope in this context. Understanding these units is crucial for interpreting the physical significance of the slope. Clear definitions of the axes and their units are essential for accurate analysis.
dannolul
Messages
4
Reaction score
0
Homework Statement
A lab was performed by taking 2 charged particles at varying distances and measuring the force between them. (Dry data, and a graph of Fe vs 1/r^2 have been given.) Assume the charges are equal of magnitude. using the slope of the line and the value of coulomb's constant constant, calculate for the charge on one particle
Relevant Equations
Fe=(kq1q2)/(r^2)
I can get the slope of the line, but then I dont know what the slope is (units or whatever it is) then i get lost from there.
 
Physics news on Phys.org
What did you plot on the ##y## axis and in what units? What did you plot on the ##x## axis and in what units? If the gradient of a ##y##-versus-##x## plot is ##\Delta y/\Delta x## then can you determine the units in this case?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top