
Which list below places the capacitors in order of increasing capacitance?
$(A)A,B,C,D$
$(B)B,A,C,D$
$(C)B,A,D,C$
$(D)A,B,D,C$
Answer
507.9k+ views
Hint: The capacitor is an electrical component that can be used to store an electric charge. Generally, a capacitor consists of two parallel plates which are placed at a distance, and in between the two parallel plates, the insulating layer or dielectric is used. We are going to find the capacitance of each capacitor given and arrange them in increasing order.
Formula used:
The capacitance of a parallel plate capacitor $C = \dfrac{{k{\varepsilon _ \circ }A}}{d}$
Where, $C$-capacitance in $\mu F$
$k$-dielectric constant
${\varepsilon _ \circ } = 8.854 \times {10^{ - 12}}F.{m^{ - 1}}$
$A$-area of parallel plates in ${m^2}$
$d$-distance between two plates in $m$
Complete step-by-step solution:
Capacitance is the measure of the ability of capacitors to store an electric charge. The unit of capacitance is $Farad(F)$.
From the given data in the diagram, the values are put into the capacitance formula is given by,
$C = \dfrac{{k{\varepsilon _ \circ }A}}{d}$
Let us consider the capacitance values of $A,B,C,D$ are ${C_A},{C_B},{C_C},{C_D}$ respectively.
The capacitance of A, ${C_A} = \dfrac{{{\varepsilon _ \circ }A}}{d}$ where $\left( {d = d} \right)$,$\left( {k = 1} \right)$ for vacuum
The capacitance of B, ${C_B} = \dfrac{{{\ varepsilon _ \circ }A}}{{2d}}$ where $\left( {k = 1,d = 2d} \right)$
The capacitance of C, ${C_C} = \dfrac{{5{\varepsilon _ \circ }A}}{d}$ where $\left( {k = 5,d = d} \right)$
The capacitance of D, ${D_D} = \dfrac{{5{\varepsilon _ \circ }A}}{{2d}}$ where $\left( {k = 5,d = 2d} \right)$
By comparing the capacitance values ${C_A},{C_B},{C_C},{D_D}$, we arrange the capacitances in increasing order. In all capacitance, we have the same terms$\left( {{\varepsilon _ \circ },A,d} \right)$ so that we are going to consider numerical values for comparison. After we have,
${C_A} = 1,{C_B} = \dfrac{1}{2},{C_C} = 5,{D_D} = \dfrac{5}{2}$
After comparing these values, the increasing order of capacitance values are
${C_B} \ll {C_A} \ll {D_D} \ll {C_C}$
Therefore the correct answer is option $(C)B,A,D,C$
Note: A capacitor is also called a condenser. The insulating materials used in capacitors are paper, mica, plastic, ceramics. capacitors are classified by the type of material used as insulators. Such classifications are Mica capacitor, Film capacitor, Paper capacitor, Electrolytic capacitor, Ceramic capacitor. Capacitance value increases with an increase in the area of parallel plates in the capacitor.
Formula used:
The capacitance of a parallel plate capacitor $C = \dfrac{{k{\varepsilon _ \circ }A}}{d}$
Where, $C$-capacitance in $\mu F$
$k$-dielectric constant
${\varepsilon _ \circ } = 8.854 \times {10^{ - 12}}F.{m^{ - 1}}$
$A$-area of parallel plates in ${m^2}$
$d$-distance between two plates in $m$
Complete step-by-step solution:
Capacitance is the measure of the ability of capacitors to store an electric charge. The unit of capacitance is $Farad(F)$.
From the given data in the diagram, the values are put into the capacitance formula is given by,
$C = \dfrac{{k{\varepsilon _ \circ }A}}{d}$
Let us consider the capacitance values of $A,B,C,D$ are ${C_A},{C_B},{C_C},{C_D}$ respectively.
The capacitance of A, ${C_A} = \dfrac{{{\varepsilon _ \circ }A}}{d}$ where $\left( {d = d} \right)$,$\left( {k = 1} \right)$ for vacuum
The capacitance of B, ${C_B} = \dfrac{{{\ varepsilon _ \circ }A}}{{2d}}$ where $\left( {k = 1,d = 2d} \right)$
The capacitance of C, ${C_C} = \dfrac{{5{\varepsilon _ \circ }A}}{d}$ where $\left( {k = 5,d = d} \right)$
The capacitance of D, ${D_D} = \dfrac{{5{\varepsilon _ \circ }A}}{{2d}}$ where $\left( {k = 5,d = 2d} \right)$
By comparing the capacitance values ${C_A},{C_B},{C_C},{D_D}$, we arrange the capacitances in increasing order. In all capacitance, we have the same terms$\left( {{\varepsilon _ \circ },A,d} \right)$ so that we are going to consider numerical values for comparison. After we have,
${C_A} = 1,{C_B} = \dfrac{1}{2},{C_C} = 5,{D_D} = \dfrac{5}{2}$
After comparing these values, the increasing order of capacitance values are
${C_B} \ll {C_A} \ll {D_D} \ll {C_C}$
Therefore the correct answer is option $(C)B,A,D,C$
Note: A capacitor is also called a condenser. The insulating materials used in capacitors are paper, mica, plastic, ceramics. capacitors are classified by the type of material used as insulators. Such classifications are Mica capacitor, Film capacitor, Paper capacitor, Electrolytic capacitor, Ceramic capacitor. Capacitance value increases with an increase in the area of parallel plates in the capacitor.
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