Why is the potential of Earth not always equal to zero?

gandharva_23
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we take the potentisl of the Earth to be equal to zero for solving some problems though the potential of Earth is not equal to zero . why is that ? or when can we do that ?
 
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Potential is relative to an arbitrary datum. You just define the zero potential surface wherever it is convenient for the calculations. So differences in potential are real, but an absolute potential does not exist. Think of it in terms of voltages -- you can define any place as zero volts, and then use your DVM to measure the voltage of other things with respect to that spot. Wherever you put the "-" lead of your DVM is automatically defined as the zero potential spot for that particular measurement.
 
we have 2 concentric spheres of radius a and b carrying charge q each . now the inner sphere is earthed . to calculate the final charge in each surface i kept the potential of the inner shell = 0 and used gauss law . am i doing a wrong thing by putting the p[otential of the inner shell = 0 ? if i m justified in doing that then that potential is zero wrt what ?
 
Zero with respect to zero. It is your reference or datum. Often, we refer to "Earth" as zero volts, but only because we use it as our voltage reference so often. The "Earth" is not at the same potential as the Sun, so we could use the voltage at the surface of the sun as our zero reference, but that wouldn't be very convenient, eh? :-)
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
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