Question 714: Electrostatics & Capacitance

Question

A parallel plate capacitor has a dielectric slab of dielectric constant $$K$$ between its plates that covers $$1/3$$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $$C$$ while that of the portion with dielectric in between is $${C_1}.$$ When the capacitor is charged, the plate area covered by the dielectric gets charge $${Q_1}$$ and the rest of the area gets charge $${Q_2}.$$ The electric field in the dielectric is $${E_1}$$ and that in the other portion is $${E_2}.$$ Choose the correct option/ options, ignoring edge effects.

Accepted Answer

Let A be area of each plate and d is the distance between the plates.
The given capacitor is equivalent to two capacitors in parallel with capacitances
$${C_1} = {{K{\varepsilon _0}(A/3)} \over d} = {{K{\varepsilon _0}A} \over {3d}}$$
$${C_2} = {{{\varepsilon _0}(2A/3)} \over d} = {{2{\varepsilon _0}A} \over {3d}}$$
$$	herefore$$ $$C = {C_1} + {C_2}$$
$$ = {{K{\varepsilon _0}A} \over {3d}} + {{2{\varepsilon _0}A} \over {3d}} = {{{\varepsilon _0}A} \over {3d}}(K + 2)$$
$$	herefore$$ $${C \over {{C_1}}} = {{K + 2} \over K}$$
Hence, option (d) is correct.
Let V be potential difference between the plates. Then
$${E_1} = {V \over d}$$ and $${E_2} = {V \over d}$$
$$	herefore$$ $${{{E_1}} \over {{E_2}}} = 1$$
Hence, option (a) is correct and option (b) is incorrect.
$${Q_1} = {C_1}V = {{K{\varepsilon _0}A} \over {3d}}V$$ and $${Q_2} = {C_2}V = {{2{\varepsilon _0}A} \over {3d}}V$$
$$	herefore$$ $${{{Q_1}} \over {{Q_2}}} = {K \over 2}$$
Hence, option (c) is incorrect.

Answer

$${{{E_1}} \over {{E_2}}} = 1$$

Answer

$${{{E_1}} \over {{E_2}}} = {1 \over K}$$

Answer

$${{{Q_1}} \over {{Q_2}}} = {3 \over K}$$

Answer

$${C \over {{C_1}}} = {{2 + K} \over K}$$

Electrostatics & Capacitance

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