Answer
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Hint: Stacking the plates alternatively will result in a parallel connection of capacitors. Since two plates are needed to make one capacitor, \[n\] number of capacitors will make \[n - 1\] capacitors.
Formula used: In this solution we will be using the following formulae;
\[{C_{eqp}} = {C_1} + {C_2} + ... + {C_n}\] where \[C\] is the equivalent capacitance of capacitors in parallel, and \[{C_{1,}}{C_2},...,{C_n}\] are the individual capacitances.
Complete Step-by-Step solution:
A parallel plate capacitor is said to be made by stacking \[n\] equally spaced plates connected alternatively. Generally, in this case, two plates of the capacitor is going to make one capacitor, and these capacitors will be in parallel with one another.
Also, since two plates make a capacitor, then \[n\] number of plates is going to make \[n - 1\] number of capacitor.
Generally, for capacitors in parallel, we have
\[{C_{eqp}} = {C_1} + {C_2} + ... + {C_n}\]where\[C\] is the equivalent capacitance of capacitors in parallel, and \[{C_{1,}}{C_2},...,{C_n}\] are the individual capacitances, and \[n\] is the number of capacitors.
Hence, for \[n - 1\] number of capacitors, we have
\[{C_{eqp}} = {C_1} + {C_2} + ... + {C_{n - 1}}\]
According to question, any two adjacent plate has a capacitance \[C\], hence we have
\[{C_{eqp}} = C + C + ... + C = \left( {n - 1} \right)C\] (because there would be \[n - 1\]\[C\]’s added together)
Hence, the resultant capacitance is \[\left( {n - 1} \right)C\]
The correct option is hence A.
Note: For clarity, we have \[n - 1\] capacitor for \[n\] plates because; as the plates of the capacitor are stacked on (or side by side of) one another, each one of the plate will act as a terminal of two capacitors, except the topmost plate and the bottommost plate, hence for example, three plate will make two capacitors because the middle plate will be a terminal to both the top and bottom
Formula used: In this solution we will be using the following formulae;
\[{C_{eqp}} = {C_1} + {C_2} + ... + {C_n}\] where \[C\] is the equivalent capacitance of capacitors in parallel, and \[{C_{1,}}{C_2},...,{C_n}\] are the individual capacitances.
Complete Step-by-Step solution:
A parallel plate capacitor is said to be made by stacking \[n\] equally spaced plates connected alternatively. Generally, in this case, two plates of the capacitor is going to make one capacitor, and these capacitors will be in parallel with one another.
Also, since two plates make a capacitor, then \[n\] number of plates is going to make \[n - 1\] number of capacitor.
Generally, for capacitors in parallel, we have
\[{C_{eqp}} = {C_1} + {C_2} + ... + {C_n}\]where\[C\] is the equivalent capacitance of capacitors in parallel, and \[{C_{1,}}{C_2},...,{C_n}\] are the individual capacitances, and \[n\] is the number of capacitors.
Hence, for \[n - 1\] number of capacitors, we have
\[{C_{eqp}} = {C_1} + {C_2} + ... + {C_{n - 1}}\]
According to question, any two adjacent plate has a capacitance \[C\], hence we have
\[{C_{eqp}} = C + C + ... + C = \left( {n - 1} \right)C\] (because there would be \[n - 1\]\[C\]’s added together)
Hence, the resultant capacitance is \[\left( {n - 1} \right)C\]
The correct option is hence A.
Note: For clarity, we have \[n - 1\] capacitor for \[n\] plates because; as the plates of the capacitor are stacked on (or side by side of) one another, each one of the plate will act as a terminal of two capacitors, except the topmost plate and the bottommost plate, hence for example, three plate will make two capacitors because the middle plate will be a terminal to both the top and bottom
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