Two identical parallel plate capacitors of same dimensions joined in series are connected to a DC source. When one of the plates of one capacitor is brought closer to the other plate:a)The voltage on the capacitor whose plates comecloser is greater than the voltage on the capacitorwhose plates are not movedb)The voltage on the capacitor whose plates comecloser is smaller than the voltage on the capacitorwhose plates are not movedc)The voltage on the two capacitors remain equald)The applied voltage is divided among the two directlyas the capacitanceCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Explanation:
When two identical parallel plate capacitors of same dimensions are joined in series and connected to a DC source, the voltage across each capacitor is equal. Let us assume that the capacitance of each capacitor is C and the voltage across each capacitor is V.

When one of the plates of one capacitor is brought closer to the other plate, the effective capacitance of that capacitor increases and becomes C'. The effective capacitance of the other capacitor decreases and becomes C''.

Effect on Voltage:
Now, the charge on each capacitor is given by Q=CV. As the voltage across each capacitor is equal, the charge on each capacitor is also equal and is given by Q=CV. When the distance between the plates of one capacitor is decreased, the effective capacitance of that capacitor increases and the charge on it also increases to Q'=C'V. At the same time, the effective capacitance of the other capacitor decreases and the charge on it decreases to Q''=C''V.

As the capacitors are in series, the total charge on the capacitors is Q=Q'+Q''=(C' + C'')V. As the charge on each capacitor is equal, we can write Q=CV.

Equating the two expressions for Q, we get CV=(C' + C'')V, which gives us C' + C''=C.

Effect on Voltage:
Now, we know that the voltage across each capacitor is given by V=Q/C. Therefore, the voltage across the capacitor whose plates are moved closer is V'=Q'/C' and the voltage across the other capacitor is V''=Q''/C''.

Substituting the values of Q', Q'' and C', C'' in the above equations, we get V'=CV/(C' + C'') and V''=CV/(C' + C'').

As C' is greater than C'', we can say that C' + C'' is greater than C'. Therefore, V' is less than V''.

Hence, the voltage on the capacitor whose plates come closer is smaller than the voltage on the capacitor whose plates are not moved. Therefore, option B is the correct answer.

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