Solving for distance between plates, diameter of a capacitor

AI Thread Summary
The discussion revolves around designing a 5.02 pF air-filled capacitor with specific parameters, including a maximum electric field of 1.40×10^4 N/C and a potential of 95 V. The calculations involve using the equations for capacitance and electric field to determine the distance between the plates and the diameter. One participant reports calculating a distance of 0.00679 m and a diameter of approximately 0.07 m, but these answers were marked incorrect by the CAPA system. There is a suggestion that the issue may relate to unit specifications or the order of inputting answers. The conversation highlights the importance of verifying calculations and understanding system requirements for accurate submissions.
PhysicsMan999
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Homework Statement



  1. A 5.02 pF air-filled capacitor with circular parallel plates is to be used in a circuit where it will be subjected to potentials of up to 95 V. The electric field between the plates is to be no greater than 1.40×104 N/C. As an electrical engineer for Live-Wire Electronics, your task is to design the capacitor. What is its size? Enter the diameter and the separation of the plates.

Homework Equations


C=kEA/d
E=V/d
A=pir^2

The Attempt at a Solution


I used E=V/d to try and find the distance, then plugged that into the capacitance equation to find the area of the plates, which I then used to find the radius of the circle, which I multiplied by 2 to get the diameter.
 
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PhysicsMan999 said:

Homework Statement



  1. A 5.02 pF air-filled capacitor with circular parallel plates is to be used in a circuit where it will be subjected to potentials of up to 95 V. The electric field between the plates is to be no greater than 1.40×104 N/C. As an electrical engineer for Live-Wire Electronics, your task is to design the capacitor. What is its size? Enter the diameter and the separation of the plates.

Homework Equations


C=kEA/d
E=V/d
A=pir^2

The Attempt at a Solution


I used E=V/d to try and find the distance, then plugged that into the capacitance equation to find the area of the plates, which I then used to find the radius of the circle, which I multiplied by 2 to get the diameter.
Just what I would have done! Congrats!
 
CAPA says my answers are wrong though! The answers I'm getting are 0.00679 for the distance between plates, and around 0.07 for the diameter..
 
PhysicsMan999 said:
CAPA says my answers are wrong though! The answers I'm getting are 0.00679 for the distance between plates, and around 0.07 for the diameter..
Did the system specify particular units to use? Is it finicky about significant figures?
 
Well, apparently I was entering them in the wrong order..even though it says nothing about the order to enter them in. Thanks!
 
PhysicsMan999 said:
CAPA says my answers are wrong though! The answers I'm getting are 0.00679 for the distance between plates, and around 0.07 for the diameter..
Well, I didn't check your math. I only endorsed your methodology! :smile:
 
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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