Where Does the Charge Go When Connected to a Grounded Capacitor?

[Nanoath]

What happens to a charged object, it can be anything, when it's connected to a capacitor, and the other side of the capacitor is grounded.
Does the charge disappear?

 

[Ofey]

No. When the charged object is connected to the capacitor(plate), the charge of the object will reduce as much as the charge increases in the capacitor (if the capacitor was uncharged before). In other words the capacitor(plate) will get the same charge as the object. (Both have either + or - charge). Since the other "side" of the capacitor is connected to the ground, an equal, yet opposite charge will "emerge" from the ground to the other plate. There is now a potential difference between the plates due to the charge differences of the plates, and thus energy is stored. Electrons cannot travel from one plate to the other because of the insulating material between them.

Charge never just "disappears" (in a local system).

 

[Ofey]

This is my guess... You might want to check with somebody whose smarter than me...

Just to make it clear. You move the charged object from the capacitor after it has given part of its charge to the plate? So they don't stay connected.

 

[Nabeshin]

No. When the charged object is connected to the capacitor(plate), the charge of the object will reduce as much as the charge increases in the capacitor (if the capacitor was uncharged before). In other words the capacitor(plate) will get the same charge as the object. (Both have either + or - charge). Since the other "side" of the capacitor is connected to the ground, an equal, yet opposite charge will "emerge" from the ground to the other plate. There is now a potential difference between the plates due to the charge differences of the plates, and thus energy is stored. Electrons cannot travel from one plate to the other because of the insulating material between them.

Charge never just "disappears" (in a local system).

Right.

 

[m.e.t.a.]

No. When the charged object is connected to the capacitor(plate), the charge of the object will reduce as much as the charge increases in the capacitor (if the capacitor was uncharged before). In other words the capacitor(plate) will get the same charge as the object.

Careful, Ofey -- the charged object and the capacitor plate will eventually reach the same electric potential, not the same charge. $C=\frac{Q}{V}$.

 

[Ofey]

Appreciate the correction