## anonymous one year ago A capacitor is charged to a potential difference of 12V. it delivers 40% of its stored energy to a lamp. what is the final potential difference across the capacitor???

1. anonymous

@Irishboy123

2. anonymous

@arindameducationusc

3. arindameducationusc

I think 12 is the charge....

4. arindameducationusc

U(Energy)=1/2*Q*V Q=charge V=potential difference (you have to find) U=40% of energy I think its this way but I am not sure, I am little weak in Electrostatics... but this should be the way....

5. IrishBoy123

$E_1 = \frac{1}{2} C V_1^2 \$ $E_2 = \frac{1}{2}C V_2^2 = 0.60 E_1$ $V_2^2 = 0.60 V_1 ^ 2$

6. anonymous

How the formula E=1/2CV^2 CAME actually...i had able to derive E=CV^2 but from where did the 1/2 came???would be very kind of you to clear me..

7. IrishBoy123

look here first https://gyazo.com/ae43ccf0b318175dcb80738cbd06ba63 any questions, post them here! basic relationship is $C = \frac{Q}{V}$ so however you derive the equation for energy, you can then manipulate it into other forms.....