# 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.

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 $VQ$.

Energy stored in a capacitor is electrical potential energy, and it is related to the charge Q and voltage V on the capacitor.

one must be careful when applying the equation for electrical potential energy.

change in potential energy(ΔPE) = Charge(q ). Change in voltage(ΔV) to a capacitor.

Remember that ΔPE is the potential energy of a charge q going through a voltage ΔV. But the capacitor starts with zero voltage and gradually comes up to its full voltage as it is charged.

The first charge placed on a capacitor experiences a change in voltage ΔV = 0,

since the capacitor has zero voltage when uncharged.

The final charge placed on a capacitor experiences ΔV = V,

since the capacitor now has its full voltage V on it.

The average voltage on the capacitor during the charging process is V/2

, and so the average voltage experienced by the full charge q is V /2.

Thus the energy stored in a capacitor, E(cap), is Ecap = QV/2, where Q is

the charge on a capacitor with a voltage V applied. (Pl. Note that the energy is not QV, but QV/2 .)

Charge and voltage are related to the capacitance C of a capacitor by Q = CV, and so

the expression for Ecap can be algebraically manipulated into three equivalent expressions:

Ecap = QV/2 = C V^2 /2 = Q^2/ 2C ,

where Q is the charge and V the voltage on a capacitor C. The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in farads.

see for details-

https://courses.lumenlearning.com/physics/chapter/19-7-energy-stored-in-capacitors/

Try calculating the charge using $V = kQ/R$. For the energy, look up “electrostatic self energy “ (it is simpler concept than it sounds) — but the basic idea is that you just integrate $Vdq$, giving you energy = $(k(Q^2))/2R$. The takeaway here is that try to avoid indirect formulas and stick to the basics — you’ll understand what you’re doing and make less mistakes.

“Where did half of the energy go?” — If you connected a battery to charge the sphere/shell using a battery, it moves a charge Q (equal to final charge on the sphere/shell) and through a potential difference V (equal to final potential of the sphere/shell). So, the total work done but the battery would be QV. Now, if you calculate energy of the sphere/shell using the method in my previous (BTW this method is correct), it’ll come out to be QV/2, i.e., half of the total work done by the battery. Where did the other half go? Lost as heat (correction — EM radiations). Maybe this is why batteries, and not capacitors, are used for storing energy.

• "Where did the other half go? Lost as heat" No, there is no energy lost as heat in the idealized situation here, where the circuit has no resistance. The energy is actually lost as electromagnetic radiation (in simple terms, you just built a radio transmitter!) - but a proper explanation of that is probably beyond the OP's current level, since it looks like the question is about "simple" electrostatics... – alephzero Aug 25 '18 at 0:55
• To analyze the "battery" circuit fully, note that the battery, the wires, and the capacitor must form a closed loop, which therefore has inductance, and therefore you have a resonant circuit, which will oscillate (and emit EM radiation) before it reaches its steady state. – alephzero Aug 25 '18 at 1:00
• @alephzero Thanks for the correction ... good to learn !! – Anurag B. Aug 25 '18 at 5:23
• @alephzero You seem to be good at this topic. I have two questions similar to this one — here and here but for some reason they didn't receive any detailed answers. Could you please take a look? – Anurag B. Aug 25 '18 at 5:36