Consider an ideal capacitor which has a length of \$\ell_1\$ between its plates. The capacitor terminals are open; they are not connected to any finite valued impedance. Its capacity is \$C_1\$ and it has an initial voltage of \$V_1\$.

What happens to the capacitor voltage if we make the gap between the plates \$\ell_2=2\ell_1\$ without changing the amount of charge on the plates?



My thoughts on this:

Increasing the gap will decrease the capacitance.

$$ C_2 = \dfrac{C_1}{2} $$

Since the amount of charge is unchanged, the new capacitor voltage will be

$$ V_2 = \dfrac{Q}{C_2} = \dfrac{Q}{\dfrac{C_1}{2}} = 2\dfrac{Q}{C_1} = 2V_1. $$

Is this true? Can we change the capacitor voltage just by moving its plates? For example, suppose that I'm wearing plastic shoes and I have some amount of charge on my body. This will naturally cause a static voltage, since my body and the ground act as capacitor plates. Now, if I climb a perfect insulator building (e.g.; a dry tree), will the static voltage on my body increase?


Question: Enabling aggregate and compress CSS files caused error on site access
Answer: I found this from http://drupal.org/node/782510

In your settings.php file, add

$conf['preprocess_css'] = CACHE_DISABLED;


Seemed to work for me on a quick test