$A$ 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......$V$

$A$ parallel plate capacitor with area $200\,cm^2$ and separation between the plates $1.5\,cm$,is connected across a battery of $emf$ $V$. If the force of attraction between the plates is $25\times10^{-6}\,N$,the value of $V$ is approximately........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$

$A$ capacitor of capacitance $C$ with a distance $d$ between its plates is connected to a battery of $V$ volts. What is the force acting between the two plates?

Four metallic plates are arranged as shown in the figure. If the distance between each plate is $d$ and the area of each plate is $A$, then the capacitance of the given system between points $A$ and $B$ is (Given $d << \sqrt{A}$):

Two parallel plate capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$ respectively. If they have the same charge and potential,and the distance between the plates of the first capacitor is $0.5 \,mm$,then the distance between the plates of the second capacitor is ........ $cm$.

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 $g = 9.8 \, m/s^2$].