# What are the potentials of a battery with respect to earth?

Assume an open-circuit voltaic cell. We know that the cell maintains a potential difference between its terminals. That is $V_+ - V_- = E$.

Now, if I connect $(-)$ to earth using the earth terminal in the mains, then $V_+ = E$ and $V_- = 0$

And If I earth $(+)$, then $V_+ = 0$ and $V_- = -E$

But what if I don't do any earthing at all? What then, is the potential of the terminals with respect to earth? (By earth I include every electrically irrelevant object, including myself)

Is $$V_+ = E/2, V_- = -E/2$$ If so, why? I am stumped as to how I should determine the potential w.r.t any outside object. In fact, I wonder what the potential of the insulating body of the cell itself is.

The voltage difference across the battery's terminals is still $1.5\,\text{V}$, but the voltage of the negative terminal may be $200\,\text{V}$, and the voltage of the positive terminal $201.5\,\text{V}$. You would not be able to measure this voltage with a multimeter, because the resistance of the multimeter (typically $1-10\,\text{M}\Omega$) would discharge this static electricity almost immediately.