As distance between two capacitor plates decreases, capacitance increases - given that the dielectric and area of the capacitor plates remain the same. So, why does this occur? As distance between two capacitor plates decreases, capacitance increases - given that the dielectric and area of the capacitor plates remain the same.
As Capacitance C = q/V, C varies with q if V remains the same (connected to a fixed potential elec source). So, with decreased distance q increases, and so C increases. Remember, that for any parallel plate capacitor V is not affected by distance, because: V = W/q (work done per unit charge in bringing it from on plate to the other) and W = F x d
The capacitors ability to store this electrical charge ( Q ) between its plates is proportional to the applied voltage, V for a capacitor of known capacitance in Farads. Note that capacitance C is ALWAYS positive and never negative. The greater the applied voltage the greater will be the charge stored on the plates of the capacitor.
Remember, that for any parallel plate capacitor V is not affected by distance, because: V = W/q (work done per unit charge in bringing it from on plate to the other) and W = F x d and F = q x E so, V = F x d /q = q x E x d/q V = E x d So, if d (distance) bet plates increases, E (electric field strength) would drecrese and V would remain the same.
The electric field between the plates of a parallel-plate capacitor To find the capacitance C, we first need to know the electric field between the plates. A real capacitor is finite in size. Thus, the electric field lines at the edge of the plates are not straight lines, and the field is not contained entirely between the plates.
Shouldn't the plates hold more charge if there are more polarised molecules in the dielectric, as the pull on the nucleus will be greater (due to all of the electrons), and thus the atom's electrons will be pulled towards the nucleus with greater force, allowing more charges on the capacitor plates? how does this increase capacitance?
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