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

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Byju's Answer

Standard XII

Physics

Breaking a Capacitor into Combinations

A parallel-pl...

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.

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Solution

The given system of the capacitor will behave as two capacitors connected in series.

Let the capacitances be C₁ and C₂.
Now,
$C_{1} = \frac{\in_{0} A k_{1}}{d_{1}} and C_{2} = \frac{\in_{0} A k_{2}}{d_{2}}$
Thus, the net capacitance is given by
$C = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{\frac{\in_{0} A k_{1}}{d_{1}} \times \frac{\in_{0} A k_{2}}{d_{2}}}{\frac{\in_{0} A k_{1}}{d_{1}} + \frac{\in_{0} A k_{2}}{d_{2}}} = \frac{\in_{0} A (k_{1} + k_{2})}{k_{1} d_{2} + k_{2} d_{1}} = \frac{(8.85 \times 10^{- 12}) \times (10^{- 2}) \times 24}{(6 \times 4 \times 10^{- 3} + 4 \times 6 \times 10^{- 3})} = 4.425 \times 10^{- 11} C = 44.25 pF$

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