Parallel plate capacitor electric charge

AI Thread Summary
In a discussion about calculating the electric charge on a parallel plate capacitor, participants explore the relationship between charge (Q), capacitance (C), and voltage (V). The total energy stored in the capacitor is expressed as U = Q^2 * d / (2 * E_o * A), with given parameters including voltage (E), dielectric constant (k), gap width (d), and plate area (A). One participant initially struggles with finding the charge from the energy formula and seeks clarification on relevant equations. They later realize the formula for surface charge density involves the electric field and surface area, but face challenges converting voltage to coulombs. Ultimately, the discussion highlights the importance of unit conversions and understanding fundamental capacitor equations.
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Consider an air filled parallel plate capacitor with plate area A and gap width d. The plate charge is Q. The total energy stored in the capacitor is given by U= Q^2 *d /2 E_o A. With the battery connected, fill the gap by a slab with the dielectric constant k. Given : E= 70 V, k= 3.8, d=0.8 mm and A = 25.6 cm^2, E_o = 8.85 x 10^-12 C^2/Nm^2, find the electric charge on the plate. Answer in units of C.
I really have no idea how to do this. I know that I have everything needed to find the total energy, but I don't really know how to get the charge from that. Can someone help?
 
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What is the equation relating Q, C and V? What is the equation relating C, epsilon, A and d?
 
I figured it out... thank you.
 
horribly similar is: given only area of each plate, a distance between the plates, and instead, k = 1 (due to a vacuum as opposed to a slab), and the voltage, how would one find surface charge density (in Coulombs/meter^2)? I think I'm only confusing myself by changing various units around (such as volts to Joules/Coulomb)

So the work I've done so far includes:
I've noticed this should be relatively easy; the formula for density is simply just magnitude of the electric field multiplied with the surface area.
First I converted surface area from cm^2 to meters^2, I'm simply stuck on changing the charge difference from volts to coulombs :(
 
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Thread 'Variable mass system : water sprayed into a moving container'

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}...
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