Question

A 800pF capacitor is charged by a 100 V battery. After sometime the battery is disconnected . The capacitor is than connected to another 800pF capacitor. What is the electrostatic energy stored ?

Answer 1

Answer:

Given,

C1 = 800pF  = $800*10^{-12}$ F

C2 = 800pF  =  $800*10^{-12}$ F

V1 = 100V

V2 = 0 [coz for the second capacitor battery is disconnected]

Common potential= V = $\frac{C1V1 + C2V2}{C1+C2}$

     

                                     = $\frac{800*10^{-12}*100+800*10^{-12}*0}{10^{-12}(800+800)}$

                                      = $\frac{80000}{1600}$

                                      = 50V

As per the question,

we need to find the energy stored ,

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

                           = $\frac{1}{2}(C1+C2)V^{2}$

                           = $\frac{1}{2}((800+800)10^{-12})*(50)^{2}$

                            = $800*10^{-12}*2500$

                           = $2*10^{6}*10^{-12} J$

                            = $2*10^{-6}J$

Thus the electrostatic energy stored is $2*10^{-6}J$

                         

 

Answer 2

$\</strong><strong>orange</strong><strong>{Answer=2'10- 6J</strong><strong>}</strong><strong>$

Capacity = 800 pF

= 8''10-10 F

V of the battery = 100 V

Hence

capacitor = CV = 8"10-8 C

Read down the notes

• If battery is dis-connected
• The capacitor is connected to another capacitor
• Same as the appenciaces, the quantities are conserved on the amount of charge on the battery

• Ratio of both capacitances

• q = 4"10-8 C on the two capacitors.

Therefore, electrostatic energy stored

consider this a C1 and C2

C1 = 800pF

= 800=10^-12800=10

10 −12 F

10 −12 FC2 = 800pF = 800=10^-12800=10

This is,

V = C1V1 + C2V2 and C1+C2

800=10^-12}

=100+800

=10^-12 and 0

10^{-12}{800+800}

800+800} 10 −12

Add them both now

800+800

800-----10

10 −12 =100+800----10

Finally 0

0 =80000 and 1600

answer here = 50V

CV^221 CV 2 C1+C2 V^{2}

V^{2} 21

(C1+C2)V 2

(800+800)

10^{-12}) and (50)^{2} 21

1 ((800+800)10 −12 )(50) 2

800--10^{-12}=

250080010 −12 =2500

2500 = 2=10^{6}

=10^{-12} J2 ''10

10 6{10 −12}=J

= 2 10^

{-6}J2=10 −6 J

Also can do like this in short method

= 2=2/2C

= 2=2/2C = =2=4=10-8

= 2=2/2C = =2=4=10-8=2/2'"8"10-10

= 2=2/2C = =2=4=10-8=2/2'"8"10-10 =

2'10-6 J

• Hence ,

The electrostatic energy stored

=2'10-6 J