homework and exercises - Electric Potential Energy of a charged conductor

My physics textbook asks (I translate):

A sphere of copper with a radius of 0.72 metres is charged with a Potential of 270,000 Volt. Find its charge and the electric energy it holds.

I found the charge correctly by first calculating the capacitance with the formula C=(1/k)·R, after which I resorted to C=Q/V, which gives me a charge of 0.0000216 Coulomb.

I then miscalculated the electric energy. I should have used the formula U=0.5·C·V^2 given somewhere in the book, but I had forgotten about its existence. Instead I used my own logic, which said that Voltage = Energy/Charge (Volt = Joule/Coulomb), which gave me precisely double the right answer (5.832 Joule instead of the right answer: 2.916 Joule).

I went over the correct formula again, and I understand it and the way it was derived via integration.

However, I am still left wondering why my original logic was wrong, and why the answer I arrived at was precisely double of what it should have been… Where did the other half go to?

Can somebody shed some light on this?

I really feel I need to understand why my logic was wrong in order not to make the same mistake again, because it was very intuitive to go down that path.

Answer

The equation $voltage=\frac{ energy}{ charge}$ or $V=\frac{E}{Q}$ should be interpreted as follows: Given a voltage of $V$, you need to invest energy $E$ to move charge $Q$ across this voltage. When you do so, you increase the energy within the system by the amount $E$. However, when you do so, the voltage itself may change. So you need to use the equation incrementally.

For the sphere, initially it is not charged, voltage (between sphere and infinity) is zero and the energy you need to invest in adding a small amount of charge, $dQ$, is also zero. The next $dQ$ would require a small amount of energy due to the newly created small voltage.The "last" $dQ$ already needs to overcome the full voltage.

Simple integration of the invested energy $\int{v(q)dq}$ would end up with half $VQ$. You can also apply some hand-waving arguments to deduce that the invested energy is an average between minimum of zero for the initial $dQ$ and $VQ$.