Question

A parallel-plate capacitor has plate area 100 cm2 and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.

Answer 1

The new dielectric capacitance = 44.25 µC

Explanation:

Given the system of the capacitor acts as two capacitors connected in series.

Considering Capacitors C1 & C2

We know Capacitance C= ∈A/d  

Now, C1 = ∈1Ak/d and C2 = ∈2Ak/d

Hence the two capacitors are connected in series, the new capacitance will be

        C = C1C2/C1 + C2

After substituting the above mentioned values in the formula we get,

=  ∈1Ak/d $\times$ ∈2Ak/d/ ∈1Ak/d + ∈2Ak/d

Simplifying the above equation

= $(8.85 \times 10-12) \times (10-12) \times 24 /(6 \times 4 \times 10-3 + 4 \times 6 \times 10-3)$

= 44.25 µC

Hence, we got that new di-electric capacitance C = 44.25 44.25 µC