How Do You Calculate the Capacitance of a Charged Capacitor?

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To calculate the capacitance of a charged capacitor, the energy stored (W) and the voltage (ΔV) are key variables. The formula W = 1/2 Q ΔV can be rearranged to find charge (Q) as Q = 2W/ΔV. For a capacitor charged at 6.10 kV storing 1140 J, the charge calculates to approximately -0.18 C. The capacitance (C) can then be determined using C = Q/ΔV, resulting in a capacitance of about -3.06E-5 F. Understanding these relationships is crucial for solving capacitor-related problems in physics.
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Homework Statement



A capacitor is in this device is charged 6.10KV and stores 1140 J of energy. What's the capacitance?

Homework Equations


W= 1/2 Q Delta V

The Attempt at a Solution


PE is W
W= 1140 J
Delta V: 6.10 KV
Q =?
C=?

Delta V = -W/Q rewrite to Q= -W/Delta V
convert kV to V
Q= -1140/6100V= -.18

then C= Q/V -.18/6100= -3.06E -5
 
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Inasmuch as you're given energy (Joules) would you happen to know a formula for finding capacitance from voltage and energy? (hint: The formula is usually written for finding energy from voltage and capacitance.)
 
if C= Q/V is what your saying, C and Q isn't given in this case there only V.
 
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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