Two parallel plate capacitors of capacitances 2 µF and 3 µF are joined in series and the combination is connected to a battery of V volts. The values of potential across the two capacitors V1 and V2 and energy stored in the two capacitors U1 and U2 respectively are related as ________.

The correct option is:

V₂/V₁ = 3/2 and U₂/U₁ = 2/3

Step-by-Step Explanation:

1. Understanding the Capacitors in Series:
We are given two capacitors with capacitances:
C1=2μF
C2=3μF
When capacitors are connected in series, the magnitude of charge (Q) on each capacitor remains the same.

2. Potential Difference Relationship:
The potential difference across a capacitor is related to its charge and capacitance by the formula:
V=QC
Let V2 be the potential difference across the 2μF capacitor (C1) and V1 be the potential difference across the 3μF capacitor (C2). This gives:
V2=QC1
V1=QC2
Taking the ratio of the two potential differences:
V2V1=C2C1=32

3. Energy Stored Relationship:
The electrostatic energy stored in a capacitor is given by the formula:
U=Q22C
Let U1 be the energy stored in the 2μF capacitor (C1) and U2 be the energy stored in the 3μF capacitor (C2). This gives:
U1=Q22C1
U2=Q22C2
Taking the ratio of the stored energies:
U2U1=C1C2=23

Thus, we establish the required relationships:
V2V1=32 and U2U1=23