mrcotton said:Homework Statement
AQA Section A Q18 Jun 11
Homework Equations
The answer should be C
E=.5QV and C=Q/V
The Attempt at a Solution
So the original Q has increased by 1.5 times If C is constant than the voltage must also increase by 1.5 times
Now the energy E is E=.5QV V has increase by 1.5 and and so has Q which is 2.25
Infact in typing this out I think I just solved it. Is this logic and method ok
The scaling factor of energy in a capacitor with change in charge is known as the capacitance. It is a measure of the amount of charge that a capacitor can store per unit of voltage.
As the charge on a capacitor increases, the capacitance also increases. This means that the capacitor can store more energy for a given voltage. The relationship between charge and capacitance is directly proportional.
The formula for capacitance is C = Q/V, where C is capacitance, Q is charge, and V is voltage. This formula can be rearranged to find the scaling factor of energy, as C = Q/V = E/V, where E is energy.
The distance between the capacitor plates, also known as the separation, has an inverse relationship with the capacitance. This means that as the distance between the plates increases, the capacitance decreases, and vice versa. This relationship is described by the formula C = εA/d, where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the separation.
Yes, the scaling factor of energy can be changed by altering the material between the plates. The permittivity of the material is a key factor in determining the capacitance. Materials with a higher permittivity, such as dielectrics, can increase the capacitance and therefore the scaling factor of energy in a capacitor with change in charge.