Aparallel-plate capacitor with plates of area 720 cm2 is charged to a potential difference v and is then disconnected from the voltage source. when the plates are moved 0.5 cm farther apart, the voltage between the plates increases by 100 v. what is the charge q on the positive plate of the capacitor?

Explanation:

Given that,

Area of the parallel plate,

Separation between two plates,

Voltage between the plates,

Let Q is the positive charge on the capacitor and is the surface charge density. It cane be given by :

The relation between electric field and the surface charge density is given by :

Also,

So,

So, the charge on the positive plate of the capacitor is . Hence, this is the required solution.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..