A parallel-plate vacuum capacitor has 7.68 JJ of energy stored in it. The separation between the plates is 3.40 . If the separation is decreased to 1.10 .

a) what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?b) What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?

B. elliptical. as in oval. 0 (==

Hello,your answer would be iron! plz mark me brainliest! hope this !

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)