The Question is from neet-physics electrostatics capacitance, and it was created by the author Mr. ANIL KUMAR.G

Question

Two identical capacitors are charged individually to voltages $V_1$ and $V_2$. They are then connected in parallel with plates of like polarity joined together.  After the system comes to equilibrium, what is the electrostatic energy stored in the system?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Capacitors in parallel, energy stored in a capacitor
 **Related Topics**: Charge conservation, voltage in parallel circuits, energy conservation
 **Prerequisites**: Understanding of capacitance, parallel circuits, energy stored in a capacitor ($U = \frac{1}{2}CV^2$)
## SOLUTION PATHWAY
**Given Information**:
  Two identical capacitors with capacitance C
  Capacitor 1 charged to $V_1$
  Capacitor 2 charged to $V_2$
  Connected in parallel with like polarities
**Method & Steps**:
 1. Calculate the initial charge on each capacitor.
  Reasoning: The charge on a capacitor is given by $Q = CV$.
  Formula/Rule: $Q_1 = CV_1$ and $Q_2 = CV_2$
 2. Determine the total charge after connecting in parallel.
  Working: Since the capacitors are connected in parallel with like polarities, the charges add up. The total charge is $Q_{total} = Q_1 + Q_2 = CV_1 + CV_2 = C(V_1 + V_2)$
  Key Point: Charge is conserved.
 3. Calculate the equivalent capacitance of the parallel combination.
  Working: When capacitors are connected in parallel, the equivalent capacitance is the sum of individual capacitances.  $C_{eq} = C + C = 2C$
 4. Find the final voltage across the parallel combination.
  Working: The total charge is divided across the equivalent capacitance.  $V_{eq} = \frac{Q_{total}}{C_{eq}} = \frac{C(V_1 + V_2)}{2C} = \frac{V_1 + V_2}{2}$
 5. Calculate the final energy stored in the system.
  Working: The energy stored in the parallel combination is given by $U = \frac{1}{2}C_{eq}V_{eq}^2 = \frac{1}{2}(2C)\left(\frac{V_1 + V_2}{2}ight)^2 = \frac{C}{4}(V_1 + V_2)^2$
  Result: $\frac{C}{4}(V_1 + V_2)^2$
  Verification: The energy should be positive and depend on both initial voltages.
 ## OPTION ANALYSIS
A ✓: $\frac{C}{4}(V_1+V_2)^2$: This is the correct result, derived above.
B :x: $\frac{C}{4}(V_1-V_2)^2$: This formula doesn't account for the addition of charges when connected in parallel.
C :x: $\frac{C}{2}(V_1^2+V_2^2)$: This is incorrect; energy is not simply additive when capacitors are connected.
D :x: $C(V_1+V_2)^2$: This omits the factor of 1/4 from the squaring of the final voltage.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly adding voltages directly instead of considering charge conservation: Remember that charge is conserved, not voltage.
  2.Forgetting to use the equivalent capacitance when calculating the final energy: The energy is stored in the parallel combination of capacitors.
 **Quick Tips**:
  1.Always start by finding the total charge and equivalent capacitance when dealing with parallel capacitor combinations.
  2.Remember the formula for energy stored in a capacitor: $U = \frac{1}{2}CV^2$

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\frac{C}{4}(V_1+V_2)^2$"

Mr. ANIL KUMAR.G · Answer

$\frac{C}{4}(V_1-V_2)^2$

Mr. ANIL KUMAR.G · Answer

$\frac{C}{2}(V_1^2+V_2^2)$

Mr. ANIL KUMAR.G · Answer

$C(V_1+V_2)^2$

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