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Magazine Chapter 4: Q8E (page 809)
A 5.00pF, parallel-plate, air-filled capacitor with circular plates are to be used in a circuit in which it will be subjected to potentials of up to 1.00 * 102 V. The electric field between the plates is to be no greater than 1.00 * 104 N/C. As a budding electrical engineer for Live-Wire Electronics, your tasks are to
(a) design the capacitor by finding what its physical dimensions and separation must be
(b) find the maximum charge these plates can hold.
Short Answer
The separation must be 4.24cm.
The maximum charge can be 500pC.
Step by step solution
About capacitance potential difference.
The capacitance C of a capacitor is the ratio of the magnitude of the charge Q on either conductor to the magnitude of the potential difference between the conductors. The capacitance depends only on the geometry of the capacitor. The greater the capacitance C of a capacitor, the greater the magnitude Q of charge on either conductor for a given potential difference V and hence the greater the amount of stored energy.
We have a bunch of variables in the above equations. So, in most problems, we are asked to get one of them by the above equation either directly by substitution or by solving the equation indirectly for this variable.
Step 2:
a)
Hence, the separation must be 4.24cm.
Calculation of charge.
b)
The capacitance is
Hence the maximum charge can be 500pC.
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