An electric circuit has two parallel plate capacitors (C1 and C2) in series. Both
capacitors have the same dielectric and inter-plate distance, but the plates in
C1 have a higher surface area compared to those in C2. Once fully charged,
which of the two capacitors will have a higher voltage drop?
C1
C2
The capacitor which is closer to the power supply
The capacitor which is further from the power supply
In a series circuit with two parallel plate capacitors C1 and C2—sharing the same dielectric and plate separation but C1 having larger plate area—the fully charged C2 experiences the higher voltage drop. This occurs because capacitance C=ϵ0κA/d makes C1 > C2, so with equal charge Q on both, voltage V=Q/C is larger across smaller C2. The total voltage divides inversely with capacitance, prioritizing the weaker link.
Capacitance Basics
Larger plate area directly boosts capacitance since C∝A, while dielectric constant κ and distance d remain identical here. In series, charge Q equals across capacitors, but voltages sum to battery voltage: V1+V2=Vtotal where V2>V1 due to C2’s lower C.
Option Breakdown
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C1: Incorrect—its higher capacitance yields lower V1=Q/C1.
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C2: Correct—the smaller area lowers capacitance, raising V2=Q/C2>V1.
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Closer to power supply: Incorrect—series position doesn’t dictate voltage split; it’s capacitance-based, not proximity.
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Further from power supply: Incorrect—same reason; voltage depends solely on C ratio, irrelevant to connection order.
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This principle applies broadly in circuits for voltage division, ensuring balanced design.
