Question

Two identical parallel plate capacitors of same dimensions are connected to a D.C. source in series When one of the plates of one capacitor is brought closer to the other plate

• A. the voltage on the capacitor whose plates came closer is greater than the voltage on the capacitor whose plates are not moved.
• B. the voltage on the capacitor whose plates came closer is smaller than the voltage on the capacitor whose plates are not moved.
• C. the voltages on the two capacitors remain equal.
• D. the applied voltage is divided equally between the two capacitors

Answer 1

Answer: (B)

the voltage on the capacitor whose plates came closer is smaller than the voltage on the capacitor whose plates are not moved.

Capacitance is given by, $C=\frac{ϵA}{d}$

where $A=$ Area of plates, $d=$ distance between the plates

If one plate is brought closed, i.e, $d$ decreases, capacitance of that capacitor will increases.

Since, $C=\frac{q}{V}$

If capacitance of capacitor is increased, Voltage of that capacitor will decreases.

So, Voltage on the capacitor whose plates came closer is smaller than the voltage on the capacitor that is not moved.