Question 514: Electrostats & Capacitance

Question

A parallel plate capacitor of capacitance ' $C$ ' is connected to a battery and charged to a potential difference ' $V$ '. Another capacitor of capacitance 3 C is similarly charged to a potential difference 3 V . The charging battery is then disconnected and capacitors are connected in parallel to each other in such a way that positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Accepted Answer

When two parallel plate capacitors are connected in parallel, their combined capacitance is the sum of their individual capacitances. HereÃ¢ÂÂs a simplified breakdown:
Capacitance in Parallel
:
Capacitor 1 has a capacitance $ C $ and is charged to a potential difference $ V $.
Capacitor 2 has a capacitance $ 3C $ and is charged to a potential difference $ 3V $.
Combined Capacitance
:
$    C_{	ext{eq}} = C + 3C = 4C    $
Net Potential Difference
:
The capacitors are connected such that the positive terminal of one is connected to the negative terminal of the other. This results in the net potential difference being the difference between the two:
$    V_{	ext{net}} = 3V - V = 2V    $
Final Energy Calculation
:
The energy stored in a capacitor is given by the formula:
$    	ext{Energy} = \frac{1}{2} C_{	ext{eq}} V_{	ext{net}}^2    $
Substituting the calculated values:
$    	ext{Energy} = \frac{1}{2} 	imes 4C 	imes (2V)^2 = 8 CV^2    $
Therefore, the final energy of the configuration is $ 8 CV^2 $.

Answer

$\frac{3}{2} \mathrm{~CV}^2$

Answer

$8 \mathrm{~CV}^2$

Answer

$\frac{13}{2} \mathrm{~CV}^2$

Answer

$18 \mathrm{~CV}^2$

Electrostats & Capacitance

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