The value of electric current on a sinking conductor

AI Thread Summary
The discussion revolves around calculating the current flowing between two plates of a capacitor immersed in kerosene. The capacitor has a specified size, voltage, and velocity of immersion, with the relative permittivity of kerosene noted. The main equation used is for capacitance, which incorporates both the depth of immersion and the permittivity of the material. Participants clarify how to handle the variable representing depth in the capacitance equation, suggesting it can be treated as a constant during differentiation. Ultimately, the focus is on deriving the current from the given parameters and equations.
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Homework Statement


A flat capacitor formed by two square plates of side 0.3 m which are 2 mm apart. Source keeps voltage 250 V on the plates. What current flows between the plates and the source if the condenser is immersing in kerosene at velocity of 5 mm / s? The relative permittivity of kerosene is 2.

Homework Equations


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The Attempt at a Solution


Here is a pic
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I=Q/t

Because of the fact that immersing cunductor can be understood as two parallel conductors the capacity of an immersed conductor (the depth x = v*t) is C =(ε0*(a-x)*a)/d + (ε0εr*a*x)/d

My question is how do I compute the time in the x = v*t? And sorry for my english.
 
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The simple answer is you don't. Just assume you know it and hope it cancels out in the end.
 
vela said:
The simple answer is you don't. Just assume you know it and hope it cancels out in the end.
But it doesn't cancel in the end because of the plus and minus in the equation
 
What equation? Remember your ultimate goal is to find the current.
 
vela said:
What equation? Remember your ultimate goal is to find the current.
I think I get it now...

I put the "v*t" instead the "x" in the equation for capacity, multiply it by the voltage (U) and then derivate it following I=dQ/dt

Or am I wrong?

Anyways thank you for your clear explanation, it helped me a lot
 
Last edited:
That's right. Or you could just leave it as ##x##, and when you differentiate with respect to time, you'll get ##v## in its place.
 
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