What does it mean to say a battery is however many volts?

[kostoglotov]

For instance, at a point 1 meter from a 1 nC charge the voltage (potential difference) from the formula V = Kq/r, is given as 9 volts, 9 J/C.

So what does it then mean to call a battery 9V, 9 J/C? Is this a battery capable of taking a 1 C charge from infinity and pushing to within 1m of a 1 nC source of some electric field?

 

[Chandra Prayaga]

It means that if you move a 1C charge from the negative terminal to the positive terminal, its potential energy increases by 9J. The battery already defines the potential difference of 9V between its positive and negative terminals, so we don't use the point at infinity as the zero potential reference.

 

[kostoglotov]

It means that if you move a 1C charge from the negative terminal to the positive terminal, its potential energy increases by 9J. The battery already defines the potential difference of 9V between its positive and negative terminals, so we don't use the point at infinity as the zero potential reference.

And so a circuit that connects those terminals is a pathway across which a charge could gain PE by moving from the negative terminal to the positive terminal? And if that circuit is more than just a (theoretically) resistance-less why, then as that charge is picking up PE, some of that PE will be doing work in the circuit before it reaches the other terminal?

 

[kostoglotov]

It means that if you move a 1C charge from the negative terminal to the positive terminal, its potential energy increases by 9J. The battery already defines the potential difference of 9V between its positive and negative terminals, so we don't use the point at infinity as the zero potential reference.

So could we think of voltage as 'height'?

 

[jbriggs444]

And so a circuit that connects those terminals is a pathway across which a charge could gain PE by moving from the negative terminal to the positive terminal?

A [positive] charge will not spontaneously move from negative terminal to positive terminal. It takes work to move it in that direction. Its gain in PE corresponds to the external work that must be supplied to get it to move.

And if that circuit is more than just a (theoretically) resistance-less why, then as that charge is picking up PE, some of that PE will be doing work in the circuit before it reaches the other terminal?

The PE that is gained by a charge does not do work as it is gained. It is the other way around. You must do work on the charge to get it to gain PE.

As a [positive] charge is released from the positive terminal and allowed to flow to the negative terminal, its potential energy is released. The PE that it loses is a source of work or heat in the circuit.

So could we think of voltage as 'height'?

Yes. That analogy works.

 

[kostoglotov]

So when Chandra Prayaga said above

"It means that if you move a 1C charge from the negative terminal to the positive terminal, its potential energy increases by 9J."

They were describing something more akin to a battery being charged than a battery being used?

 

[jbriggs444]

So when Chandra Prayaga said above

"It means that if you move a 1C charge from the negative terminal to the positive terminal, its potential energy increases by 9J."

They were describing something more akin to a battery being charged than a battery being used?

Yes.

 

[Chandra Prayaga]

Yes.

Thanks, jbriggs444. Yes. The electric potential energy increases by 9J. That energy needs to be provided by the hand or whatever moves the positive charge. In the motion of any charge q from point A to point B, any number of forces could be acting on the charge, but the change in electric potential energy is U_B - U_A = q(V_B - V_A), where V stands for electric potential and U stands for electric potential energy.