Hello Forum,
Now lets consider a charge conductor carrying charge Q_initial=+1C and planet Earth.
Planet Earth can be considered to be a huge spherical conductor of radius 6,371 kilometers that carries a negative surface charge density of about -1 nC/square meter and capacitance ~ 710 microfarad. the Earth's surface is 510.1 trillion m² which implies that its total negative surface charge is 5.7*10^5 C. Lots of negative charge!
When we ground a charged object, the object and Earth becomes a single equipotential conductor that has the same electric potential V_equilibrium at every point. Charge will move around from high potential to low potential and distributes itself until a new equilibrium potential is set. The equipotential system (object+Earth) will carry a net negative charge since (Q_object+Q_earth ) ~ Q_earth. The highest negative surface charge density, but the smallest amount of charge, will be found on the "discharged" object which in reality has some charge and does not really get discharged at all.
In fact, regardless of the sign and amount of its initial charge, a charged object that is grounded will eventually carry a small amount of negative charge and will not become completely discharged.
Capacitance is C=Q/V and for Earth C is only 710 microfarad. But the earth charge Q_earth (5.7*10^5 C ) is so huge that V_earth=Q/C does not change much at all when it is connect with a charged object, i.e. when Earth receives or gives away a little amount of charge. After grounding, the equilibrium electric potential that the grounded object+Earth have is very close (equal for all practical purposes) to the initial Earth's potential V_earth=(9*10^9)*(5.7*10^5 C )/(6,371*10^3) (relative to infinity where V is set to zero).
Since it is potential difference is the quantity that matters, we often set V_earth=0 instead of using a point at infinity...
Are my observations correct? A grounded object, no matter its amount and sign of charge, does not get completely discharged and the earth potential is not affected (much) and remains constant...
thanks
Spacerat
Now lets consider a charge conductor carrying charge Q_initial=+1C and planet Earth.
Planet Earth can be considered to be a huge spherical conductor of radius 6,371 kilometers that carries a negative surface charge density of about -1 nC/square meter and capacitance ~ 710 microfarad. the Earth's surface is 510.1 trillion m² which implies that its total negative surface charge is 5.7*10^5 C. Lots of negative charge!
When we ground a charged object, the object and Earth becomes a single equipotential conductor that has the same electric potential V_equilibrium at every point. Charge will move around from high potential to low potential and distributes itself until a new equilibrium potential is set. The equipotential system (object+Earth) will carry a net negative charge since (Q_object+Q_earth ) ~ Q_earth. The highest negative surface charge density, but the smallest amount of charge, will be found on the "discharged" object which in reality has some charge and does not really get discharged at all.
In fact, regardless of the sign and amount of its initial charge, a charged object that is grounded will eventually carry a small amount of negative charge and will not become completely discharged.
Capacitance is C=Q/V and for Earth C is only 710 microfarad. But the earth charge Q_earth (5.7*10^5 C ) is so huge that V_earth=Q/C does not change much at all when it is connect with a charged object, i.e. when Earth receives or gives away a little amount of charge. After grounding, the equilibrium electric potential that the grounded object+Earth have is very close (equal for all practical purposes) to the initial Earth's potential V_earth=(9*10^9)*(5.7*10^5 C )/(6,371*10^3) (relative to infinity where V is set to zero).
Since it is potential difference is the quantity that matters, we often set V_earth=0 instead of using a point at infinity...
Are my observations correct? A grounded object, no matter its amount and sign of charge, does not get completely discharged and the earth potential is not affected (much) and remains constant...
thanks
Spacerat