Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

A parallel-plate capacitor is located horizontally, so that one of its plates is submerged into liquid while the other is over its surface (figure). The permittivity of the liquid is equal to $ε$ , its density is equal to $ρ$ . To what height will the level of the liquid in the capacitor rise after its plates get a charge of surface density $σ$ ?

Accepted Answer

Suppose the charge density on the upper face of the capacitor is $+ σ$ and the face of liquid just opposite to the upper plate is $σ$ . The electric field inside the capacitor, where there is no liquid, is $E = \frac{σ}{ε_{0}}$ . If permittivity of the liquid is $ε$ , then electric field inside the liquid must be equal to $\frac{E}{ε}$ .

As the liquid will also develop equal and opposite charge density on its two sides $(σ')$ , the electric field due to charge density (say $E'$ ) on the liquid will be opposite to that of electric field due to capacitor plates ( $E$ ). The net electric field inside the liquid will be the difference in electric fields, i.e., $E - E'$ .
So,

$E - E' = \frac{E}{ε} \frac{σ}{ε_{0}} - \frac{σ'}{ε_{0}} = \frac{σ}{ε_{0} ε} σ' = \frac{ε - 1}{ε} σ$

We can see one extra electrostatic attraction force between the upper plate of capacitor ( $+ σ$ ) and the upper surface of the liquid ( $- σ'$ ). Due to this electrostatic attraction force, the liquid will rise extra height (suppose $h$ ) from the liquid outside the capacitor.
So, Electrostatic attraction force $=$ Weight of the liquid rise

Charge on the surface of the liquid $\times$ Field due to the upper capacitor plate $= ρ V g$

$σ' A \times \frac{σ}{2 ε_{0}} = ρ A h g \frac{ε - 1}{ε} σ \times \frac{σ}{2 ε_{0}} = ρ h g h = \frac{(ε - 1) σ^{2}}{2 ε_{0} ε ρ g}$

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