you have it reversed. capacitance increases with decrease in distance of plates.
Answer: Capacitance is unaffected by frequency; it does not change. Details: Capacitance is unaffected by frequency. In a capacitor, what increases with Frequency is Admittance (analogus to Conductance) . The capacitive Reactance is inversely proportional to Frequency. Therefore, when Frequency is increased, current flow may increase.
The main role of dielectrics in capacitors is to increase the value of capacitance of the capacitor.
A: Any additional capacitor added in parallel will effectively increase to total capacitance by that value. Note that additional capacitor added must have the same voltage rating as the other
(a) Charge Will increase (b) Potential difference will stay the same (c) Capacitance will increase (d) Stored energy will decrease
any capacitance is given by equation C = (epsilon * A/ d) where d is distance between two plates, thus as d reduces C increases. Now, in depletion region as we increase reverse bias, the depletion region width increases. Now consider depletion region as a parallel plate capacitor, with positive charges on n side and negative charges on p side. Thus, as reverse bias increases, d of junction capacitance increases thus capacitance reduces. On other hand, as reverse bias reduces, d of junction capacitance reduces, thus capacitance increases. -Amey Churi
Answer: Capacitance is unaffected by frequency; it does not change. Details: Capacitance is unaffected by frequency. In a capacitor, what increases with Frequency is Admittance (analogus to Conductance) . The capacitive Reactance is inversely proportional to Frequency. Therefore, when Frequency is increased, current flow may increase.
Not much if it is a fixed capaitor. If it is a variable capacitor, rotating the shaft to mesh the two sets of plates more closely will increase the capacitance.
The main role of dielectrics in capacitors is to increase the value of capacitance of the capacitor.
Ripples will increase if capacitance is decreased.
A: Any additional capacitor added in parallel will effectively increase to total capacitance by that value. Note that additional capacitor added must have the same voltage rating as the other
A: Capacitance vary directly with the area applied the bigger the area the bigger the capacitor. There is another element that will increase the size that is the rating of the capacitor voltage since more material will have to be used to insulate the plates
The capacitance will remain the same. However, the energy stored (Ie. number of electrons displaced) will increase.
in a series RC circuit phase angle is directly proportional to the capacitance
Definitely not possible. Capacitance is given by an expression C = epsilon x A / d Since charge is not present the capacitance cannot be increased or decreased by the charge placed
(a) Charge Will increase (b) Potential difference will stay the same (c) Capacitance will increase (d) Stored energy will decrease
That will depend on the dielectric. There will be two main effects - any change in the permeativity of the dielectric, and thermal expansion which will increase the distance between the plates. There will also be an change (probably an increase), in leakage current through the dielectric. Any change is very likely to be small or insignificant - I have worked in a factory making capacitors and temperatures were very variable, not controlled, for measurements of capacitance.
capacitance also increase