Capacitor's voltage

Question

Imagine I have two plates that form a capacitor, so the magnitude of the charge in each plate is, let's say, $Q$. So, the force, in respect to the distance $d$, is gonna be: $$F(d) = k\frac{Q^2}{d^2}$$ So, in order to calculate the energy stored in the capacitor, since it's equal the work needed to pull a plate next to the other, I can calculate the energy $E$ by: $$E = F(d)\cdot d$$ But since $F$ changes with respect to $d$, then I have to integrate $F$, and then multiply by $d$, so:

$$E = \int F(d)dd = \int k\frac{Q^2}{d^2} = kQ^2\int\frac{1}{d^2}dd = -kQ^2\frac{1}{d}$$

So: $$E = -kQ^2\frac{1}{d} =-k\frac{Q^2}{d}$$

I also know that voltage is the same as energy per charge, so: $$V = \frac{E}{Q} = \frac{-k\frac{Q^2}{d}}{Q} = -k\frac{Q}{d}$$

Is it right?

I don't know, because the capacitance $C$ is $$C = \frac{Q}{V}$$ or $$Q = CV$$ And $E = QV$ Then if I substitute $Q$ inside my energy formula, I get:

$$E = -k\frac{CV^2}{d}$$ but the right should be:

$$E = \frac{CV^2}{2}$$

Can somebody help me?

Answer 1

The equation of $F(d)$ you cite ($F(d)=k\frac{Q^2}{d^2}$)is only for two point charges and not for the force in case of two charged plates. If you have two large (theoretically infinite) charged plates, each having surface charge density $\sigma=Q/A$, then the force on any one plate is $\frac{\sigma}{2\epsilon_o}Q$ and is not dependent on the distance in the case of very large plates.
If you find the potential energy $U$ by integrating this from $0$ to $d$:-
$$\int_0^d \frac{\sigma}{2\epsilon_0}Qdx$$ you will get $\frac{\sigma}{2\epsilon_0}Qd$ which then can be rearranged using $C=\frac{\epsilon_0 A}{d}$ to get $U=\frac{Q^2}{2C}$ or $U=\frac 12 CV^2$.

Answer 2

The energy stored in an electric field is given by $$ W= \frac{\epsilon_0}{2}∫dr^3 E⃗^ 2(r⃗ )$$