$A$ parallel plate capacitor has $1\,\mu F$ capacitance. One of its two plates is given $+2\,\mu C$ charge and the other plate,$+4\,\mu C$ charge. The potential difference developed across the capacitor is......$V$

The lower plate of a parallel plate capacitor is supported on a rigid rod. The upper plate is suspended from one end of a balance. The two plates are joined together by a thin wire and subsequently disconnected. The balance is then counterpoised. Now a voltage $V = 5000 \, V$ is applied between the plates. The distance between the plates is $d = 5 \, mm$ and the area of each plate is $A = 100 \, cm^2$. Find the additional mass $m$ (in grams) placed to maintain balance. [All elements other than plates are massless and nonconducting. Take $\epsilon_0 = 8.85 \times 10^{-12} \, F/m$ and $g = 9.8 \, m/s^2$].

The potential difference between two plates of a parallel plate capacitor is $2V$. As shown in the figure,electrons are placed at point $P$ and $Q$. So,

Assertion : The total charge stored in a capacitor is zero.
Reason : The field just outside the capacitor is $\frac{\sigma }{{{\varepsilon _0}}}$. ( $\sigma $ is the charge density).

Six metallic plates,each with a surface area $A$ on one side,are placed at a distance $d$ from each other. The alternate plates are connected to points $P$ and $Q$ as shown in the figure. The capacitance of the system is:

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles,then