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A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in capacitors .If another capacitor of 6pF connected in with it with same battery connected across the combination. Find the charge stored and potential difference across the capacitance.?

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Answer 1

E=1/2CV^2`

`E=1/2×12×10^-12×(50)^2`

`E=1.5×10^-8 J`

`C_(eqv)=(C_1C_2)/(C_1+C_2)`

`=(12×6)/(12+6)`

`=72/18`

`=4×10^-12 F`

`Q=C_(eqv)V`

`=4×10^-12×50`

`=200×10^-12 C`

Charge stored in both capacitor is `200×10^-12 C`.

Voltage across 12 pF capacitor is `=Q/C= (200×10^-12)/(12×10^-12)=200/12=16.67 V`

Voltage across 12 pF capacitor is`=Q/C= (200×10^-12)/(6×10^-12)=200/6=33.33 V`

Answer 2

Given,

C1 = 12pF  = $12*10^{-12}F$

[To convert IpF in F , we multiply the given value by $10^{-12}$ )

As per to the question,

Electrostatic Energy = $\frac{1}{2}CV^{2}$

                          = ½ * 12* $10^{-12}$ * 50 * 50

                         = 15000 * $10^{-12}$

                        = 1.5 * $10^{-8}$ J

Here we know,

C2 = 6pF  =  $6*10^{-12}F$

C1 and C2 are in series.  (Given)

1/C = 1/12 + 1/6

1/C = ¼

C = 4pF =  $4*10^{-12}F$

Charge stored = CV

                 = 4 * $10^{-12}$* 50

                 = 2 * $10^{-10}$ C

Additional data :

Capacitor :

A capacitor is an arrangement which can store sufficient quantity of charge.

Suppose , on giving a charge Q to a conductor , the electric potential of the conductor becomes V.

Then the capacitance of the conductor is

$c=\frac{Q}{V}$

Capacitor in series :

If several capacitors are connected in series, then the reciprocal of the capacitance of the equivalent capacitor is equal to the sum of the reciprocals of the capacitances of the individual capacitor.

$\frac{1}{C}=\frac{1}{C1}+\frac{1}{C2}+\frac{1}{C3}$

Energy stored in a charged conductor :

The total amount of work in charging the capacitor  is stored up in the capacitor in the form of electric potential energy.

$U=\frac{1}{2}CV^{2}$