What happens of charges on other side of plates of capacitor

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When a capacitor is connected to a battery, positive charge accumulates on one plate while negative charge accumulates on the opposite plate. The charges induced on the opposite sides of each plate do not remain there; they are transferred to the battery as electrons move. For calculating the electric field between the plates, only the charges on the plates are considered, as they are responsible for the electric field and voltage. The structure of capacitors, often multilayered or cylindrical, enhances capacitance and minimizes inductance, allowing for a contained electrostatic field. Overall, the charges delivered from the battery to the capacitor are crucial for understanding the capacitor's function and electric field calculations.
Dexter Neutron
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When a capacitor is connected to battery then positive charge is induced on one plate and a negative charge is induced on another plate.Now what happens to the charges that are induced on opposite sides of each plate i.e. on plate with positive charge a negative charge would be induced on opposite side and same for other plate.
Why these charges are not taken into account for calculating electric field between the plates?
 
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Those charges go to the battery. They don't stay on the plates.
 
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As each electron is moved by battery emf to one plate, another electron is repelled off the other plate. As already posted.
Dexter Neutron said:
Why these charges are not taken into account for calculating electric field between the plates?

They are. The field results from the charge [and voltage] : V/d = E and Q=CV
 
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Dexter Neutron said:
When a capacitor is connected to battery then positive charge is induced on one plate and a negative charge is induced on another plate.Now what happens to the charges that are induced on opposite sides of each plate i.e. on plate with positive charge a negative charge would be induced on opposite side and same for other plate.
Why these charges are not taken into account for calculating electric field between the plates?
Although the capacitor icon may be two parallel plates, in practice these are metal foils and will be multilayered or rolled up into a cylinder so that the reverse side of one plate comes up against the other (but separated by an additional film of dielectric) so it now forms double the capacitance. In the case of a cylindrical form, the spiral edges of the foil are then crimped together so it more resembles a host of concentric cylinders rather than one endlessly-long spiral. This produces the same capacitance while minimizing its inductance and resistance, so making it much closer to the ideal than would otherwise be the case.

In addition, the cylindrical construction means that its electrostatic field can be wholly contained within the device and isn't going to induce interfering signals in nearby circuits.
 
when a parallel plate capacitor is connected to a closed ciruit and according to the conventional agreement of current direction, charges move from the negative terminal of the battery to the plate of the capacitor connected to it (as a result there would the excess electrons on this plate) then it repels equal amount of electrons from the plate connected to the positive terminal of the battery(the electrons move to the positive terminal of the battery) so in calculations the charge delivered the battery to the capacitor is considered. hope it helped
 
It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.
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