MisterP said:Homework Statement
Capacitor has a charge of 0.008C. Capacity is 2 μF. Need to calculate energy of electric field. Answer: 32J
Homework Equations
E = q2/2*C
The Attempt at a Solution
This seems right equation, but instead of 32J I get 16J.. any ideas?
A capacitor's electric field is the force field created by the separation of positive and negative charges on the plates of the capacitor. It is responsible for storing electrical energy in the capacitor.
The energy in a capacitor's electric field can be calculated using the formula E = 1/2CV^2, where C is the capacitance of the capacitor and V is the voltage across the capacitor.
The energy in a capacitor's electric field is directly proportional to its capacitance. This means that as the capacitance increases, the energy stored in the electric field also increases.
The energy in a capacitor's electric field is directly proportional to the square of its voltage. This means that as the voltage increases, the energy stored in the electric field increases exponentially.
The energy in a capacitor's electric field can be affected by the capacitance, voltage, and the type of dielectric material used in the capacitor. Additionally, any external factors such as temperature and humidity can also impact the energy stored in the electric field.