The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

For a parallel plate capacitor having plates with finite area,the electric field lines bend outward at the edges. This effect is called . . . . . . .

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

$A$ thin aluminum sheet of negligible thickness is placed between the plates of a parallel plate capacitor. The capacitance of the capacitor:

The capacitance of a parallel plate capacitor is $12\,\mu F$. If the distance between the plates is doubled and the area is halved,then the new capacitance will be.........$\mu F$.

$A$ metal slab of thickness $b = d/2$ is inserted between the plates of a parallel plate capacitor of plate separation $d$. What is the ratio of the new capacitance to the original capacitance?