Say you're out for a spacewalk on the ISS one day, and you see an unknown object floating next to you (ignoring the probability that it somehow matched your orbit perfectly and didn't just plunge straight into Earth's atmosphere). Let's say a bowling-ball sized metallic sphere.

Given the relative nature of voltage/potential difference, assuming this thing hasn't come into contact with anything else recently and ignoring any other ways a mysterious floating object could kill you, is it possible that the potential difference between you and the object could be so great that contact with it would instantly/almost instantly kill you? Would the situation be any different in an atmosphere, or with a bigger object?

  • $\begingroup$ The largest object around would probably be the ISS - and they're grounded to ISS.I don't know how but it's not a wire. And they're in an atmosphere - it's called the ionosphere - which is a plasma created by the Sun. $\endgroup$ Nov 27, 2019 at 0:12
  • $\begingroup$ Food for thought: perfect vacuum has high dielectric strength (sadly no infinity) making it a good insulator... so it is still prone to breakdown😎 $\endgroup$
    – user6760
    Nov 27, 2019 at 0:42

2 Answers 2


The short answer it is highly unlikely it will cause lethal electric shock. Here is the reason why.

Lethal electric shock requires current of sufficient magnitude and duration to flow through your heart to cause ventricular fibrillation or cardiac arrest. This means there has to be two contacts on the body with the source of current in order to have a current path through the heart. Examples are current paths from chest to back, hand to foot, and hand to hand.

Now, for this unknown object to potentially cause lethal electric shock there are three minimum requirements:

  1. There has to be a potential difference, or voltage, between two points on the object that is potentially capable of driving the necessary current through the heart.

  2. The object has to be capable of delivering the necessary current, which depends on the source impedance and available electrical energy.

  3. The shape/dimensions of the object having the potential for causing shock has to be such that two simultaneous contacts with it and the body per 1 is possible.

Based upon the above it seems highly unlikely, to me at least, that a spherical metal ball could cause a fatal electric shock. Primarily because it’s unlikely to meet the first requirement above that there will be a potential difference between two points on a metal (conducting) sphere.

I realize that it wouldn't be able to deliver current in the usual sense, but what about charge equalization? Could that produce any significant effect, or am I misunderstanding how that works?

The phenomenon you are describing is called electrostatic induction. The static electric field associated with the sphere would not penetrate electrically conductive objects such as the human body. Instead it would induce a surface electric charge on the body. If physical contact is made the charges should combine and the two body become electrically neutral. Generally, the physical sensation of this migration of charge to the surface of the skin is limited to that associated with its interaction with body hair.

Bottom line: I believe that any internal currents generated in the body during electrostatic induction should be brief and occur at or near the surface of the body and not involve critical organs, notably the heart. Just like the electrostatic discharges (ESD) we experience here on earth, I believe it should be harmless to persons.

Hope this helps.

  • $\begingroup$ I realize that it wouldn't be able to deliver current in the usual sense, but what about charge equalization? Could that produce any significant effect, or am I misunderstanding how that works? $\endgroup$ Nov 27, 2019 at 19:05
  • $\begingroup$ @IsaacMiddlemiss When you say charge equalization, are you referring to the possibility of a static discharge from the sphere to the person? $\endgroup$
    – Bob D
    Nov 27, 2019 at 19:12
  • $\begingroup$ Yes, exactly (Had to google to make sure that's what I was talking about) $\endgroup$ Nov 27, 2019 at 21:04
  • $\begingroup$ @IsaacMiddlemiss OK, I will update my answer to respond. $\endgroup$
    – Bob D
    Nov 27, 2019 at 21:36
  • $\begingroup$ @IsaacMiddlemiss I have updated my answer. Hope it helps. $\endgroup$
    – Bob D
    Nov 27, 2019 at 22:28

Putting your hands on opposite terminals of a large, high voltage capacitor will do it. They are dangerous.

  • $\begingroup$ A metal sphere is likely to be an equipotential surface. How does your capacitor analogy apply here? $\endgroup$
    – Bob D
    Nov 27, 2019 at 20:07
  • $\begingroup$ I did not mean an analogy. I meant an actual capacitor. A charged metal sphere would have to have an extremely high voltage to hold enough charge and energy for a dangerous shock. Likely the high voltage would cause it to discharge. Though neutral overall, a capacitor can hold a much larger charge on its plates at a high enough voltage to kill. It would not be an equipotential surface. You would have to touch both terminals. $\endgroup$
    – mmesser314
    Nov 27, 2019 at 23:05
  • $\begingroup$ If the charged metal sphere is one plate of a capacitor, where is the other plate? $\endgroup$
    – Bob D
    Nov 28, 2019 at 1:53
  • $\begingroup$ I may be answering a different question than the one you asked. "Lets say a bowling ball sized capacitor." Such a capacitor has two terminals. $\endgroup$
    – mmesser314
    Nov 28, 2019 at 4:31
  • $\begingroup$ Just to be clear, I have no problem with the statement you made in your answer. I just wasn't sure how it applied to the OP example of a charged metallic sphere. If in fact there were a charged "bowling ball sized capacitor" it could cause, as you say, an electric shock if the voltage and capacitance were sufficient and the two plates were accessible. Sorry if there was a misunderstanding. $\endgroup$
    – Bob D
    Nov 29, 2019 at 2:16

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