What's the difference between electrostatic voltage and normal voltage, like the battery's voltage.
How to calculate the charge on a charged plate if we knew its electrostatic voltage?
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1$\begingroup$ Hi Selim. Can you point me a reason on what's preventing you from nearing the fact that both are exactly the same..? ;-) $\endgroup$– Waffle's Crazy PeanutCommented Apr 20, 2013 at 12:51
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2 Answers
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Voltage is voltage. Electrostatic voltage is still just voltage.
Materials that can store electrostatic charges are insulators (or are insulated by something), so the charge cannot leak away. Because of this they can build up a high voltage, with relatively few electrons (charge). A conducting material on the other hand needs a power source to keep the voltage up, as the charge continuously drains away.
Voltage is always measured with respect to something else. It's the difference in potential between two points in a circuit. You can treat the charged plate as one plate of a capacitor, with the other plate being ground. You can then calculate the voltage from the charge and the capacitance with this formula:
$$V=Q/C$$
Where V is the voltage, Q is the charge and C is the capacitance (in Farads). As you can see, the smaller the capacitance, the larger the voltage.
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A more "straight" answer to your question, is:
$Q = V \cdot C, = V \cdot \frac{\kappa A}{d}$. $Q$ = charge; $V$ = voltage, $C$ = capacitance; $\kappa$ = permittivity constant; $A$ = surface area holding the charge; $d$ = distance between the surfaces holding the charge.
The formula changes, depending on the shape of the surfaces - parallel plates, cylinders, spheres, etc. and any dielectric that might be in between the surfaces.
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1$\begingroup$ Hi Guill. Welcome to Physics.SE. This site uses an unique TeX markup style called MathJax. This markup is very useful for understanding math equations and parameters. Please have a look here for an intro or our FAQ for more info. For example,
$\theta$results $\theta$,$\omega$inserts $\omega$, etc. It's quite interesting. You can revise your post if you can ;-) $\endgroup$ Commented May 21, 2013 at 1:50