The force of attraction between the plates of a parallel plate capacitor is

The intensity of the electric field at a point between the plates of a charged parallel plate capacitor:

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

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).

$A$ capacitor is made of a flat plate of area $A$ and a second plate having a stair-like structure as shown in the figure. If the area of each stair is $\frac{A}{3}$ and the height of each step is $d$,the capacitance of the arrangement is:

$A$ parallel plate capacitor has an electric field of $10^5 \, V/m$ between the plates. If the charge on the capacitor plate is $1 \, \mu C$,the force on each capacitor plate is ...... $N$.