5
$\begingroup$

I have a copper conductor. For a while, I apply a voltage of $12kV$ DC from a source. After removing the source, will the conductor stay charged from the source if is not earthed? Will it discharge when I connect the conductor to earth?

$\endgroup$
1
  • $\begingroup$ Step 1: get a gold-leaf electroscope. Step 2: charge the electroscope. Step 3: done. $\endgroup$ Feb 20 '17 at 18:33
2
$\begingroup$

A voltage source acts as an electron pump. Suppose we take a battery as an example, then in the battery a chemical reaction pumps electrons from the cathode to the anode. The cathode becomes depleted in electrons and becomes positive while the anode acquires an excess of electrons and becomes negative. As soon as enough electrons have been pumped for the resulting potential difference to equal the reaction potential the reaction stops.

The point of all this is that suppose we attach some conductors to the battery in the way you describe. We'll make them spheres for simplicity:

Battery

If a charge $Q$ is transferred then the voltage on conductor A is:

$$ V_A = \frac{Q}{C_A}$$

where $C_A$ is the capacitance of conductor A, and likewise:

$$ V_B = -\frac{Q}{C_B}$$

At equilibrium the potential difference $V_A-V_B$ will be equal to the battery voltage so:

$$ V_\text{batt} = Q\left(\frac{1}{C_A} + \frac{1}{C_B}\right) $$

So given the voltage of the battery and the capacitance of the two conductors we can calculate how much charge is transferred. This will be non-zero, so the conductors will become charged and if you disconnect them from the battery they will keep that charge. So the answer to your question:

After removing the source, will the conductor stay charged from the source if is not earthed?

is yes.

But you need the capacitances on both sides of the voltage source to calculate how much charge is transferred, and you only say what is connected to one side of your 12kV voltage source. If the other side is connected to earth the capacitance is effectively infinite i.e. $1/C_\text{earth}=0$ in our equation above. However if the other side of your 12kV source is not connected to anything the capacitance is close to zero so $1/C\approx\infty$ and plugging this into our equation we gind $Q\approx 0$ i.e. a negligable charge is transferred.

So while it's generally true that in the situation you describe the conductor will end up charged, you haven't given us enough information to say what that charge is.

$\endgroup$
2
  • $\begingroup$ To be more specific. Im working in cable factory where we are rewinding conductors with insulation on it. We are rewind from one reel to another and check the insulation with spark tester. It is the machine that apply 12kV DC voltage on insulated conductor and when the insulation is broken, the machine give us a signal. Im trying to find out working principle of this machine and only logic explanation for me is potential difference. If the insulation is broken, the conductor with the DC source is potentialy equals and till the conductor is not grounded, it creates safety risk. Am I right? $\endgroup$
    – trenccan
    Feb 20 '17 at 18:36
  • $\begingroup$ @trenccan: In general the amount of charge transferred will be very small, and it's likely to leak away rapidly. I think it's extremely unlikely you'd even notice any shock from residual charge left on the wire, let alone find it dangerous. $\endgroup$ Feb 20 '17 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.