# The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2 microfarad. The separation is reduced to half and it is filled with the substance dielectric of value 2.8 . The new capacity of the capacitor is

Samyak Jain
333 Points
3 years ago
Capacitance (or capacity) of parallel plate capacitor is C = $\varepsilon$0A/d, where A is the area of either plate and d is the distance between them.
Let C1 = 2 $\mu$F = 2 x 10–6 F , d1 = 0.4 cm = 4 x 10–3 m
So, 2 x 10–6 = $\varepsilon$0A/4 x 10–3  $\dpi{100} \Rightarrow$ $\varepsilon$0A = 8 x 10–9 Fm
When distance between plates is reduced to half, d2 = 2 x 10–3 m and
new capacitance C2 = $\varepsilon$0A/d2 = 8 x 10–9 / 2 x 10–3 = 4 x 10–6 F
Dielectric substance is also filled. In this case, new capacitance will become dielectric times initial capacitance.
C’ = k C2 = 2.8 x 4 x 10–6 = 11.2 x 10–6 F or 1.12 x 10–5, which is the required capacity of the capacitor.