Calculating Rate of Change of Electric Field in a Circular Capacitor

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SUMMARY

The discussion focuses on calculating the rate of change of the electric field in a circular parallel plate capacitor with a radius of 0.1m and a separation of 0.10 mm. The potential difference across the capacitor increases at a rate of 10 V/μs, leading to a calculated rate of change of the electric field of 1 x 10^11 V/m. Participants explored the relationships between electric field (E), charge (Q), and voltage (V), utilizing equations such as ⌡E.dA = Q/ε and Φ = Q/ε to derive the correct answer.

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  • Familiarity with the equations of electrostatics, specifically ⌡E.dA = Q/ε
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  • Basic calculus concepts, particularly derivatives
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dekoi
A circular parallel plate capacitor has plates of radius 0.1m; the plates are separated by 0.10 mm. A current is supplied to charge the capacitor. While charging, the potential difference across the capacitor increases by 10 V/μs.

What is the magnitude of the rate of change of the electric field between the capacitor plates? Answer= 1 x 10^11 V/m.

What exactly is the "rate of change of the electric field"? dE/dt? dΦ/dt? Φ?
I have tried different methods using all three of those and have yet to receive the correct answer.

I am using: ⌡E.dA = Q/ε for dE/dt
and Φ= Q/ε for dΦ/dt and Φ
 
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I get the correct answer. What is the equation for electric field, in terms of voltage and distance? You are given dV/dt, and the separation distance...

Just watch your units and exponents.
 
Wow! Thanks for the quick reply, I'm in urgent need right now.


Thank you for the information, as it helped me get the right answer.
 

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