
A parallel plate capacitor with air as an electric has capacitance $C$. A slab of dielectric constant $K$ , having the same thickness as the separation between the plates is introduced so as to fill one- fourth of the capacitor as shown in the figure. The new capacitance will be:

A) $\left( {K + 3} \right)\dfrac{C}{4}$
B) $\left( {K + 2} \right)\dfrac{C}{4}$
C) $\left( {K + 1} \right)\dfrac{C}{4}$
D) $\dfrac{{KC}}{4}$
Answer
136.5k+ views
Hint: Using the area of the space filled by the dielectrics, calculate it for the air. Substitute both the areas in the capacitance of the dielectrics and the air formula. Add the capacitance of the air and dielectrics to find the answer.
Useful formula:
(1) The formula of the capacitance is given by
$C = \dfrac{{ \in A}}{d}$
(2) The formula of the capacitance of the air is given by
$c = \dfrac{{{ \in _ \circ }A}}{d}$
Complete step by step solution:
It is given that the
The dielectric constant of the slab is $K$
The area filled by the dielectric is $\dfrac{1}{4}$
Using the formula of dielectric initially,
$c = \dfrac{{{ \in _ \circ }A}}{d}$
The air fills the rest of the space of the dielectric.
The capacitance of the air finally is $c = \dfrac{{{ \in _0}\left( {1 - \dfrac{1}{4}} \right)A}}{d}$
By simplifying the above equation.
$c = \dfrac{{{ \in _0}\left( {\dfrac{3}{4}} \right)A}}{d}$
Substituting the formula of the capacitance of the air initially in the final capacitance.
$c = \dfrac{{3C}}{4}$
The area of the dielectric material is $\dfrac{1}{4}$ of the capacitor.
Using the formula (1) to find the capacitance of the dielectric.
$C = \dfrac{{ \in A}}{d}$
Substituting the value of the $A$ of the dielectrics.
$C = \dfrac{{ \in \dfrac{A}{4}}}{d}$
Substituting the capacitance of dielectrics in the above equation
$C = \dfrac{{K{ \in _0}\dfrac{A}{4}}}{d}$
$C = \dfrac{{KA}}{d}$
The capacitance of the parallel plate capacitor is the sum of the capacitance of the air and the capacitance of the dielectrics.
${C_{net}} = c + C$
${C_{net}} = \dfrac{{3C}}{4} + \dfrac{{KC}}{4}$
${C_{net}} = \dfrac{{\left( {K + 3} \right)C}}{4}$
Thus the option (A) is correct.
Note: The dielectrics is the poor conductor of electricity. Hence they are used as the insulator between the two plates in the parallel plate conductor. It is used for the process of electric polarization by causing the alignment of the positive charges and the electrons.
Useful formula:
(1) The formula of the capacitance is given by
$C = \dfrac{{ \in A}}{d}$
(2) The formula of the capacitance of the air is given by
$c = \dfrac{{{ \in _ \circ }A}}{d}$
Complete step by step solution:
It is given that the
The dielectric constant of the slab is $K$
The area filled by the dielectric is $\dfrac{1}{4}$
Using the formula of dielectric initially,
$c = \dfrac{{{ \in _ \circ }A}}{d}$
The air fills the rest of the space of the dielectric.
The capacitance of the air finally is $c = \dfrac{{{ \in _0}\left( {1 - \dfrac{1}{4}} \right)A}}{d}$
By simplifying the above equation.
$c = \dfrac{{{ \in _0}\left( {\dfrac{3}{4}} \right)A}}{d}$
Substituting the formula of the capacitance of the air initially in the final capacitance.
$c = \dfrac{{3C}}{4}$
The area of the dielectric material is $\dfrac{1}{4}$ of the capacitor.
Using the formula (1) to find the capacitance of the dielectric.
$C = \dfrac{{ \in A}}{d}$
Substituting the value of the $A$ of the dielectrics.
$C = \dfrac{{ \in \dfrac{A}{4}}}{d}$
Substituting the capacitance of dielectrics in the above equation
$C = \dfrac{{K{ \in _0}\dfrac{A}{4}}}{d}$
$C = \dfrac{{KA}}{d}$
The capacitance of the parallel plate capacitor is the sum of the capacitance of the air and the capacitance of the dielectrics.
${C_{net}} = c + C$
${C_{net}} = \dfrac{{3C}}{4} + \dfrac{{KC}}{4}$
${C_{net}} = \dfrac{{\left( {K + 3} \right)C}}{4}$
Thus the option (A) is correct.
Note: The dielectrics is the poor conductor of electricity. Hence they are used as the insulator between the two plates in the parallel plate conductor. It is used for the process of electric polarization by causing the alignment of the positive charges and the electrons.
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