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It's stated that if we increase the amount of charge on a capacitor the voltage across the capacitor also increases. And if we increase the surface area of a capacitor we will have a greater capacitance, because the bigger the surface the more charge that can be stored in the capacitor. How is this possible? I mean if we increase amount of charge the voltage across the capacitor also increases. So the capacitance should remain the same? And how about distance between plates of capacitance? How does it affect the capacity?

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  • $\begingroup$ Re, "...if we increase the amount of charge on a capacitor the voltage across the capacitor also increases." That is true **IF** you increase the charge without changing the dimensions of the capacitor (i.e., you increase the charge by forcing current "through" the capacitor.) $\endgroup$ Feb 3, 2020 at 17:07

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Suppose that you have a parallel plate capacitor with plate area $A$ and separation $d$.

With a potential difference across the plates of $V$ the charge stored on the capacitor is $Q$.

Now consider another capacitor with the same dimensions and the same potential difference across the plates and hence the same charge stored $Q$

If you join the plates of the two capacitors together then you now have a capacitor with a plate area $2A$, a potential difference $V$ across the plates and a total charge of $2Q$.

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Increasing the capacitance will increase the charge, but voltage will still remain constant (assuming the capacitor is still plugged in to your voltage source). There are good capacitor simulations online that can help you visualize this. Like this one.

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"if we increase the surface area of a capacitor we will have a greater capacitance, because the bigger the surface the more charge that can be stored in the capacitor. How is this possible? I mean if we increase amount of charge the voltage across the capacitor also increases. "

It is actually wrong in the case of capacitors. You know that we take $A>>d$ and this affects the area concept.

The potential between plates can be expressed as $V$ where $V = E d$

And the Electric field doesn't remains dependent on area of plates instead its charge density. Do you know the equation $E=\sigma /2 \epsilon$ for Electric field by a uniformly charged plate? Since the area is too much large than distance between plates $E$ only depends upon charge denisty and permittivity

I hope it solves the problem.

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