Why does the area of the plates affect the capacitance? Lets say I have a parallel plate capacitor with a charge of 10C and a potential difference of 5V. By the definition $C=Q/V$, the capacitance is 2 farads. Now if I increase the area of the plates, the charge definitely doesn't change. By the definition $V=kq/r$, the voltage doesn't change either. So why does the capacitance increase if we increase the surface area?
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1$\begingroup$ That's not the right formula for the voltage of a plate. $\endgroup$– MichaelCommented Apr 23, 2013 at 3:47
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$\begingroup$ I didn't see any formula for voltage on that site. What is the correct formula then? $\endgroup$– OviCommented Apr 23, 2013 at 3:57
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$\begingroup$ Just multiply the field by the distance between the plates. If you put it all together you'll find that the voltage is proprtional to $Qd/A$ where $A$ is the area of the plates and $d$ is the distance between them. $\endgroup$– MichaelCommented Apr 23, 2013 at 4:10
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$\begingroup$ Very simple thoughts will show You that the capacity is proportional to the area. So, instead of making an obviously wrong statement, You should think about where the mistake in Your reasoning is, or maybe ask for help to find that mistake. $\endgroup$– GeorgCommented Apr 23, 2013 at 14:08
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3 Answers
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The capacitance is the ratio of charge on the plates over the voltage applied. $$C = \frac{Q}{V} \Leftrightarrow Q = C \cdot V$$ The calculation you show determines the capacitance from measured voltage and charge on the plates. You basically know the result you want and determine the size of the capacitor you need.
A larger capacitor, with a larger capacity, will hold a bigger charge at the same voltage. Doubling the area will double the capacitance (in case of a plate capacitor), so for 4 farads of capacity you get $$Q = C \cdot V = 4 F \cdot 5 V = 20 C$$
The pysics works as follows: The voltage is a driving force, pushing electrons through the wires an onto the plates of the capacitor (or sucking them off on the positive pole), until the mutual repulsion of the electrons leads to a balance of foces. If you have a larger plate, the charge can distribute over a larger area, there is less "pileup" and therefore a smaller "pushback force". This is why, with larger plates, you get a bigger charge into your capacitor with the same voltage.
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The capacitance $C=\frac{Q}{U}$ is always constant for any types of capacitors. as $Q$ is increased $U$ also increase so that the fraction $C$ remains constant.
capacitor is fixed for particular size of capacitor. greater the size of capacitor, greater will be its capacitance.
Capacitance is analogous to the capacitance of water tank at our home. larger the size of tank, larger will be its capacitance despite the presence of water in tank or empty. An empty tank or water filled tank has same capacitance. similarly, whether the capacitor is charged or not charged its capacitance is always fixed. Capacitance is independent of charge and voltage raised due to this charge.
answered May 2, 2013 at 10:49
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You might be able to understand the physics if we do the problem in reverse. Assume you have a 1 farad capacitor and a 5 volt source. The maximum charge would be 5C. In other words, if the capacitor is unable to hold the charge, it won't! You can't arbitrarily decide how much charge a given capacitor can hold, this is determined by the physical characteristics of the capacitor, namely the area of the plates and the separation between them. This is given by C = kA/d, where A is the plates area and d their separation. From this, you should be able to see that if you double the area, you double the capacitance, if you double the separation, you cut it in half.
