Consider two capacitances of capacity C1 and C2 which are connected in series and have potential difference V. What is the potential difference across C1?

The correct option is V(C₂/C₁ + C₂).

To understand why this is the correct answer, let us break down the physical concepts and mathematical steps involved in a series combination of capacitors.

Step 1: Understanding Charge in Series Connection
When two capacitors with capacitances _C1 and _C2 are connected in series across a total potential difference _V, the charge _Q stored on each capacitor is the same.
Therefore, we have:
_{Q1 = Q2 = Q}

Step 2: Relate Charge, Capacitance, and Potential Difference
The potential difference across any capacitor is given by the formula:
$V=\frac{Q}{C}$
Let _V1 be the potential difference across the first capacitor _C1, and _V2 be the potential difference across the second capacitor _C2. We can write:
$V_1=\frac{Q}{C_1}$
and
$V_2=\frac{Q}{C_2}$

Step 3: Relate Individual Voltages to Total Voltage
The total potential difference _V across the series combination is the sum of the individual potential differences:
_{V = V1 + V2}
Substituting the expressions for _V1 and _V2 into this equation gives:
$V=\frac{Q}{C_1}+\frac{Q}{C_2}$
Factoring out the common charge _Q:
$V=Q\left(\frac{1}{C_1}+\frac{1}{C_2}\right)$
Finding a common denominator inside the parentheses:
$V=Q\left(\frac{C_1+C_2}{C_1{C}_2}\right)$

Step 4: Solve for Charge Q
Rearranging the equation to solve for the charge _Q:
$Q=V\left(\frac{C_1{C}_2}{C_1+C_2}\right)$

Step 5: Calculate Potential Difference across C₁
Now, substitute the value of _Q back into the expression for _V1:
$V_1=\frac{Q}{C_1}$
$V_1=\frac{1}{C_1}·V\left(\frac{C_1{C}_2}{C_1+C_2}\right)$
Simplifying the expression by canceling out _C1 from the numerator and denominator:
$V_1=V\left(\frac{C_2}{C_1+C_2}\right)$

Thus, the potential difference across the capacitor _C1 is indeed _{V(C₂/C₁ + C₂)}.