There is no -Q induced on the negative plate since according to the problem the capacitor has been disconnected from the circuit; there is no path for charge to move onto or off of the negative plate. So there is only a +Q charge on the positive plate.EddiePhys said:Charge +Q is given and hence -Q is induced on the negative plate.
The formula ##V=Q/C## assumes equal and opposite charge on both plates. Since only one plate was charged here, it is only half of this value ##Q/2C##.EddiePhys said:The potential difference should be V+Q/C
But the answer is V+Q/2C. Why?
Charge given externally to a capacitor refers to the process of adding or removing electric charge to or from a capacitor through an external source, such as a battery or power supply.
The amount of charge stored in a capacitor determines its capacitance, which is the ability to store electric charge. The higher the charge, the higher the capacitance, and the greater the potential energy stored in the capacitor.
Yes, the charge on a capacitor can be changed by altering its physical components, such as the distance between its plates or the type of dielectric material used. These changes can affect the capacitance and therefore the amount of charge the capacitor can hold.
When a capacitor is connected to a circuit, the charge on the capacitor begins to flow through the circuit, creating a current. As the charge is used up, the voltage across the capacitor decreases until it reaches equilibrium with the voltage of the rest of the circuit.
The charge on a capacitor can be calculated using the formula Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor. Additionally, the charge can be determined by integrating the current over time, as Q = ∫I(t)dt.