Can the energy expression 1 2 cv 2 of a capacitance be replaced by the variable v with an equation

Your question is: Can the terminal voltage be converted into current in the formula w=1 2cv 2 stored in a capacitor?

A: Communication should be possible; Depending on the situation, DC will generally say that no current is allowed to pass through.

Because: AC should be OK, because, there is a split second current in a cycle, and the current through the capacitor is not zero.

If the current flowing through the capacitor is generated by an alternating voltage or an alternating current source, the current will periodically change direction, and the AC current will take turns charging the two poles of the capacitor, and the charge at the two poles of the capacitor will change periodically, so that the current through the capacitor is not zero in a cycle, except for the moment when the current changes from positive to negative (or from negative to positive). For this reason, capacitors are generally considered to allow alternating current to pass through. “

2.DC depends on the situation, because the external electric field separates the positive and negative charges of the dielectric slightly.

Depending on the properties of the dielectric, the external electric field will slightly separate the positive and negative charges of the dielectric or arrange the orientation of the dielectric molecules according to the direction of the external electric field, which will form a surface charge and its corresponding electric field on the surface of the dielectric. ”

In addition: -- I think you can also use the human body static electricity phenomenon in winter (is it considered DC?). Because the human body is equivalent to a capacitor, but when the human body is charged with a lot of static electricity, this electricity does not do work on the human body, such as making the human body produce Joule heating. Therefore, in this case, V cannot be replaced with i.

When the electricity of the high-voltage wire is transmitted to the earth, the car tires on the other side of the road are blackened, and the people in the car are stunned, indicating that there is a current at this time. Therefore, it should be possible to convert to i in the case of communication.

i＝q/t

v=it c (v=0 at t=0).

1/2)(c*v^2)=(1/2)(c*i^2*t^2/c^2)=(1/2)(i^2*t^2/c）=(i^2*t^2)/(2c)..t=0, v=0).

Note: If you have learned integrals, the expression of the integral formula is the most rigorous, the expression i with the integral formula can change with t, and the above formula i cannot change with t.

Well, interesting.

In fact, this is the case, just like a porter, if the charge of q is carried according to the voltage of u all the time, the work done is w=uq

However, the capacitance is different, and the u of the capacitance is not constant, but a variable quantity.

In terms of the process of change, only dq=cdu, that is, a small amount of q, moved du

So w= cudu

On the whole, in the process of capacitors moving charges, there is no U at the beginning, and then it gets bigger and bigger, and finally USome moved small, some large, on average 1 2u

Here's the difference between ++i and i++.

i = i++;is to copy first and then add itself, which is equivalent to i = 0;

i= ++i;The result is i =1

i++ and ++i are two different operations, the former is that i becomes i+1 after the result is i, and the latter is to change i to i+1 and then i+1 as the result output.

There are a little more programs to look at.

There are formulas to estimate.

Capacitors reserve energy.

w＝1/2cv²

v This is small.

The capacitance of a capacitor is usually expressed in terms of F, NF, PF i.e. microfar, nanofar, picofar, and kilowatt-hour is the unit of energy.

1.The energy stored in a capacitor is proportional to the square of the voltage across the capacitor! Therefore, the v in the equation e= refers to the voltage across the capacitor, not to the amount of change in its voltage.

2.A rise in voltage indicates an increase in the amount of energy stored by the capacitor, which is proportional to the square difference of the voltage, not the square of the difference. That is, to use the formula you gave later.

3.The book you read is calculated using the previous formula, and it is estimated that you are reading it wrong, and people who can write books will never make this kind of mistake that pigs do not make. Or it may be said in the book that "the potentials of the two plates of the capacitor are v0 and v1 respectively, then the voltage at both ends of the capacitor is v=v1-v0, and the stored electric field energy is." ”

U=W Q means that when the voltage is U and the amount of electricity flowing through is Q, the work of the circuit is W, and U is a fixed value in the process. The capacitance U rises gradually during the charging process (U is also 0 when q is 0), and U is a variable in the process, so the energy formula of the capacitor cannot be derived by U=W Q.

It seems that here you should use the capacitor potential energy formula. Instead of using your voltage formula.