Why can voltage be negative? Watch

Yatayyat
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So I've been looking at the charge and discharge of a capacitor, and how voltage changes with time when a capacitors charges up or discharges.

Let's say I take up a a basic circuit set up, where there is a cell, parallel plate capacitor and fixed resistor all connected to each other; with a switch too.

When the switch is closed,I know that V_r decreases from EMF of cell to zero but conversely V_c increases from zero to EMF of cell. That means that at any moment of time 'V_c + V_r = EMF' since all the pd's of the cell must add up to EMF.

But there is a different side to discharging a capacitor. If I were to discharge the same capacitor when it has been fully charged up, I need to take out the cell and reconnect the circuit again. But how exactly does the voltage of the fixed resistor and capacitor change during that discharge.

I've been told that:

"V_c decreases from minus EMF to zero and V_r decreases from EMF to zero, therefore at any given time there is a relationship always that V_c = minus V_r."

I don't get the above statement that is quoted. Could someone please explain why it has that relationship and why pd can also be negative in this case?

Thanks!
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Eimmanuel
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(Original post by Yatayyat)
So I've been looking at the charge and discharge of a capacitor, and how voltage changes with time when a capacitors charges up or discharges.

Let's say I take up a a basic circuit set up, where there is a cell, parallel plate capacitor and fixed resistor all connected to each other; with a switch too.

When the switch is closed,I know that V_r decreases from EMF of cell to zero but conversely V_c increases from zero to EMF of cell. That means that at any moment of time 'V_c + V_r = EMF' since all the pd's of the cell must add up to EMF.

But there is a different side to discharging a capacitor. If I were to discharge the same capacitor when it has been fully charged up, I need to take out the cell and reconnect the circuit again. But how exactly does the voltage of the fixed resistor and capacitor change during that discharge.

I've been told that:

"V_c decreases from minus EMF to zero and V_r decreases from EMF to zero, therefore at any given time there is a relationship always that V_c = minus V_r."

I don't get the above statement that is quoted. Could someone please explain why it has that relationship and why pd can also be negative in this case?

Thanks!
During charging,


  \mathcal{E} = V_C + V_R



You know the above equation.

During discharging,   \mathcal{E}  = 0, so



0 = VC + VR
VC = -VR


Remember that the capacitor charge is given by



 q(t) = Q_0 e^{-t/RC}



And the discharging current in the circuit is



 I(t) = -\dfrac{dq}{dt}= \dfrac{Q_0}{RC} e^{-t/RC}


Note that   \mathcal{E}  = Q_0/C.



 I(t) =\dfrac{\mathcal{E}}{R} e^{-t/RC}


If we multiply both sides by R,



 V_R (t) = \mathcal{E} e^{-t/RC}


Since VC = -VR,


 V_C (t) = -\mathcal{E} e^{-t/RC}


This explains the following:
(Original post by Yatayyat)
…I've been told that:
"V_c decreases from minus EMF to zero and V_r decreases from EMF to zero, therefore at any given time there is a relationship always that V_c = minus V_r."
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Yatayyat
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(Original post by Eimmanuel)
During charging,




  \mathcal{E} = V_C + V_R





You know the above equation.

During discharging,   \mathcal{E}  = 0, so





0 = VC + VR
VC = -VR




Remember that the capacitor charge is given by





 q(t) = Q_0 e^{-t/RC}





And the discharging current in the circuit is





 I(t) = -\dfrac{dq}{dt}= \dfrac{Q_0}{RC} e^{-t/RC}




Note that   \mathcal{E}  = Q_0/C.





 I(t) =\dfrac{\mathcal{E}}{R} e^{-t/RC}




If we multiply both sides by R,





 V_R (t) = \mathcal{E} e^{-t/RC}




Since VC = -VR,




 V_C (t) = -\mathcal{E} e^{-t/RC}




This explains the following:
Thanks a lot! I get how the relationship between V_c and V_r works now from seeing the equations that you shown me.

But I'm still bit concerned why voltage can be negative in the first place. Isn't it always meant to be positive value? I get why it can be negative in order for the EMF to be zero. But I always thought that voltage is positive given that I know it to be a scalar quantity.
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Eimmanuel
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(Original post by Yatayyat)
Thanks a lot! I get how the relationship between V_c and V_r works now from seeing the equations that you shown me.

But I'm still bit concerned why voltage can be negative in the first place. Isn't it always meant to be positive value? I get why it can be negative in order for the EMF to be zero. But I always thought that voltage is positive given that I know it to be a scalar quantity.
Who says that scalar quantity must always be positive?

Try to measure the p.d. across a resistor in a closed circuit using a multimeter by reversing the polarity and see what to the value of the voltage in the multimeter.
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Yatayyat
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(Original post by Eimmanuel)
Who says that scalar quantity must always be positive?

Try to measure the p.d. across a resistor in a closed circuit using a multimeter by reversing the polarity and see what to the value of the voltage in the multimeter.
So is that if you change the polarity in a circuit only then would the voltage switch from being positive or negative, given off by the reading on the multimeter?

In which case a capacitor is able to do to the circuit when it does discharge.
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Eimmanuel
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(Original post by Yatayyat)
So is that if you change the polarity in a circuit only then would the voltage switch from being positive or negative, given off by the reading on the multimeter?
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Yatayyat
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(Original post by Eimmanuel)
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Oh okay, makes more sense from the image. Thanks again for help! It was really helpful
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anosmianAcrimony
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I'm no electrician, but would I be right in saying that voltage is force available to move charged particles, and that polarity in voltage describes which direction around a circuit, or through a wire, that force is available to move the charge?
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Yatayyat
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(Original post by anosmianAcrimony)
I'm no electrician, but would I be right in saying that voltage is force available to move charged particles, and that polarity in voltage describes which direction around a circuit, or through a wire, that force is available to move the charge?
But is it not that P.d. is the work done (hence energy transferred) to move one coulomb of charge from one point to another? Not the force available to each charge unit.

In other words voltage is how much joules of energy a coulomb of charge gets e.g. 3 volts gives 3 joules of energy to each coulomb.
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anosmianAcrimony
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(Original post by Yatayyat)
But is it not that P.d. is the work done (hence energy transferred) to move one coulomb of charge from one point to another? Not the force available to each charge unit.

In other words voltage is how much joules of energy a coulomb of charge gets e.g. 3 volts gives 3 joules of energy to each coulomb.
Having looked into it more - yes, voltage can be described as the energy available to move charge through a wire or circuit, measured in terms of energy per unit charge. It was the idea of electromotive force that threw me off...

I think my explanation of why voltage can be negative is still basically accurate, though. If a voltage is positive, that says energy is available to move charge one direction around a circuit; if it's negative, that means energy is available to move charge in the opposite direction - right?
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Eimmanuel
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(Original post by Yatayyat)
Thanks a lot! I get how the relationship between V_c and V_r works now from seeing the equations that you shown me.

But I'm still bit concerned why voltage can be negative in the first place. Isn't it always meant to be positive value? I get why it can be negative in order for the EMF to be zero. But I always thought that voltage is positive given that I know it to be a scalar quantity.
Let me add something in regard to the scalar quantity stuff. I am not sure why it seems quite a numbers of A level students have the thinking that scalar quantity is positive. Thinking about work done which is also a scalar quantity but it can be negative. Hope the following link can help to reinforce your understanding of scalar quantity.
https://www.quora.com/Can-a-scalar-quantity-be-negative
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Eimmanuel
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(Original post by anosmianAcrimony)
....I think my explanation of why voltage can be negative is still basically accurate, though. If a voltage is positive, that says energy is available to move charge one direction around a circuit; if it's negative, that means energy is available to move charge in the opposite direction - right?
Not really sure what you are trying to say.
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Athematica
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I think people assume scalar is equivalent to modal which obviously isn’t the case. Temperature isn’t moving in any particular direction. The particles with energy may be but then we only describe the relative average energy in a particular system compared to a fixed standard. That we have positive and negative scalar quantities is relatively arbitrary. We could have easily have made all temperature values positive but water’s freezing point is a convenient reference
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anosmianAcrimony
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(Original post by Eimmanuel)
Not really sure what you are trying to say.
Well, I can't explain it any more clearly than that. Try harder.
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anosmianAcrimony
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(Original post by Athematica)
I think people assume scalar is equivalent to modal which obviously isn’t the case. Temperature isn’t moving in any particular direction. The particles with energy may be but then we only describe the relative average energy in a particular system compared to a fixed standard. That we have positive and negative scalar quantities is relatively arbitrary. We could have easily have made all temperature values positive but water’s freezing point is a convenient reference
Yeah, but voltage's polarity (whether it's negative or positive) actually has a meaning to it - it's not just an arbitrary point on a scale. If we measure the voltage across two contacts with a voltmeter and then do so again, but attach the voltmeter electrodes to the contacts the other way around, the second reading will be -1 x the first reading. As I understand it, voltage polarity describes in which direction there is energy available to move charge. People call voltage a scalar, but I like to think of it as a vector in a 1D environment - it's either pushing charge one direction around a circuit, or in the opposite direction.
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Athematica
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(Original post by anosmianAcrimony)
Yeah, but voltage's polarity (whether it's negative or positive) actually has a meaning to it - it's not just an arbitrary point on a scale. If we measure the voltage across two contacts with a voltmeter and then do so again, but attach the voltmeter electrodes to the contacts the other way around, the second reading will be -1 x the first reading. As I understand it, voltage polarity describes in which direction there is energy available to move charge. People call voltage a scalar, but I like to think of it as a vector in a 1D environment - it's either pushing charge one direction around a circuit, or in the opposite direction.
It’s the difference in electrical potential rather than the measure of direction in the actual electric field.
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Athematica
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That’s because the work done is independent of the way in which is travels
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Eimmanuel
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(Original post by anosmianAcrimony)
Having looked into it more - yes, voltage can be described as the energy available to move charge through a wire or circuit, measured in terms of energy per unit charge. It was the idea of electromotive force that threw me off...

I think my explanation of why voltage can be negative is still basically accurate, though. If a voltage is positive, that says energy is available to move charge one direction around a circuit; if it's negative, that means energy is available to move charge in the opposite direction - right?
(Original post by anosmianAcrimony)
Well, I can't explain it any more clearly than that. Try harder.
(Original post by anosmianAcrimony)
Yeah, but voltage's polarity (whether it's negative or positive) actually has a meaning to it - it's not just an arbitrary point on a scale. If we measure the voltage across two contacts with a voltmeter and then do so again, but attach the voltmeter electrodes to the contacts the other way around, the second reading will be -1 x the first reading. As I understand it, voltage polarity describes in which direction there is energy available to move charge. People call voltage a scalar, but I like to think of it as a vector in a 1D environment - it's either pushing charge one direction around a circuit, or in the opposite direction.

Before I discuss about the meaning of negative voltage or p.d., I would set up some definitions such that we are on the same ground to define potential difference.



Assume that some source charge distribution set up an electric field  \vec{E} .
A charge q is placed in the electric field, so the charge will experience an electric force  q\vec{E} .
Note that this electric force is conservative.

As the charge q is displaced by an infinitesimal amount  d\vec{s}, the electric potential energy of the charge–field system is changed by an amount

 dU=-{{W}^{\text{electric}}}=-q\vec{E}\cdot d\vec{s} .


For a finite displacement of the charge from some point A in space to some other point B, the change in electric potential energy of the system is

 \Delta U = U_B - U_A =-q\int\limits_{A}^{B}{\vec{E}\cdo  t d\vec{s}} .

Because the electric force is conservative, the integral does not depend on the path taken from A to B.

For a given position of the charge in the field, the charge–field system has a potential energy U relative to the configuration of the system that is defined as U = 0. We define the electric potential energy of a system of charges to be zero when the charges are all infinitely separated from each other.

Using the conventional reference configuration, UA = 0, the work done by the electric forces between particles during the move in from infinity

 U_B - U_A =-{{W}^{\text{electric}}} .

Since UA = 0, the final potential energy UB of the system can simply be denoted as U.

 U =-{{W}^{\text{electric}}} .
This defines the electric potential energy of the system.

Dividing the electric potential energy by the charge gives a physical quantity that depends only on the source charge distribution and has a value at every point in an electric field. This quantity is called the electric potential V:

 V = \dfrac{U}{q}


The potential difference  V_B - V_A between two points A and B in an electric field is defined as the change in electric potential energy of the system when a charge q is moved between the points divided by the charge:

 V_B - V_A = \dfrac{\Delta U}{q} =-\int\limits_{A}^{B}{\vec{E}\cdot d\vec{s}}


Let’s now consider the situation in which an external agent moves the charge q in the electric field.
Assume that the external agent moves the charge q from A to B without changing the kinetic energy of the charge, the external agent performs work that changes the electric potential energy of the system:

 \Delta U = U_B - U_A = ={{W}^{\text{external}}}


So the work done by an external agent in moving a charge q through an electric field at constant velocity is

Wexternal = q ΔV

Now we are in the position to discuss the polarity of p.d. between 2 points and then apply to circuit elements.
Instead of writing p.d. as ΔV, I would write it more explicitly.

If A is the initial point and B is the final point, the p.d. between A and B is

VA→B = VBVA

When VA→B = VBVA < 0, there is a loss of electric potential energy in moving the charge q from A to B. This implies that the work is done by the field. There is a simple way to think about it. Say we have a positive point charge Q1 fixed at the origin. A free to move positive point charge q is brought near to it. How would the point charge q move naturally? The point charge q would move away from Q1 to minimize the repulsive force and electric potential energy.

When VA→B = VBVA > 0, there is a gain of electric potential energy in moving the charge q from A to B. This implies that the negative work is done by the field or an external agent is required to do the work.

Be careful in using the above interpretation to interpret circuit voltage or p.d. I would use conventional current instead of electron current. Let says that the current is flowing from A to B, what this means is that the electric potential at B is lower than that at A or VB < VA.

When we use a voltmeter to measure the p.d. between A and B, we need to place the red coloured lead at point A and the black coloured lead at B in order for us to get a positive voltage reading. If we switch the positions for the red and black coloured lead, we would get a negative voltage reading.

Does it contradict with the above interpretation of positive and negative potential difference? Not really.
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ThatOldGuy
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Voltage can be negative because it had bad parenting. It happens. Voltage has Daddy issues and that contributes to its tendency to oscillate between positive and negative. It's a little bipolar that way.
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anosmianAcrimony
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(Original post by Eimmanuel)
Before I discuss about the meaning of negative voltage or p.d., I would set up some definitions such that we are on the same ground to define potential difference.



Assume that some source charge distribution set up an electric field  \vec{E} .
A charge q is placed in the electric field, so the charge will experience an electric force  q\vec{E} .
Note that this electric force is conservative.

As the charge q is displaced by an infinitesimal amount  d\vec{s}, the electric potential energy of the charge–field system is changed by an amount

 dU=-{{W}^{\text{electric}}}=-q\vec{E}\cdot d\vec{s} .


For a finite displacement of the charge from some point A in space to some other point B, the change in electric potential energy of the system is

 \Delta U = U_B - U_A =-q\int\limits_{A}^{B}{\vec{E}\cdo  t d\vec{s}} .

Because the electric force is conservative, the integral does not depend on the path taken from A to B.

For a given position of the charge in the field, the charge–field system has a potential energy U relative to the configuration of the system that is defined as U = 0. We define the electric potential energy of a system of charges to be zero when the charges are all infinitely separated from each other.

Using the conventional reference configuration, UA = 0, the work done by the electric forces between particles during the move in from infinity

 U_B - U_A =-{{W}^{\text{electric}}} .

Since UA = 0, the final potential energy UB of the system can simply be denoted as U.

 U =-{{W}^{\text{electric}}} .
This defines the electric potential energy of the system.

Dividing the electric potential energy by the charge gives a physical quantity that depends only on the source charge distribution and has a value at every point in an electric field. This quantity is called the electric potential V:

 V = \dfrac{U}{q}


The potential difference  V_B - V_A between two points A and B in an electric field is defined as the change in electric potential energy of the system when a charge q is moved between the points divided by the charge:

 V_B - V_A = \dfrac{\Delta U}{q} =-\int\limits_{A}^{B}{\vec{E}\cdot d\vec{s}}


Let’s now consider the situation in which an external agent moves the charge q in the electric field.
Assume that the external agent moves the charge q from A to B without changing the kinetic energy of the charge, the external agent performs work that changes the electric potential energy of the system:

 \Delta U = U_B - U_A = ={{W}^{\text{external}}}


So the work done by an external agent in moving a charge q through an electric field at constant velocity is

Wexternal = q ΔV

Now we are in the position to discuss the polarity of p.d. between 2 points and then apply to circuit elements.
Instead of writing p.d. as ΔV, I would write it more explicitly.

If A is the initial point and B is the final point, the p.d. between A and B is

VA→B = VBVA

When VA→B = VBVA < 0, there is a loss of electric potential energy in moving the charge q from A to B. This implies that the work is done by the field. There is a simple way to think about it. Say we have a positive point charge Q1 fixed at the origin. A free to move positive point charge q is brought near to it. How would the point charge q move naturally? The point charge q would move away from Q1 to minimize the repulsive force and electric potential energy.

When VA→B = VBVA > 0, there is a gain of electric potential energy in moving the charge q from A to B. This implies that the negative work is done by the field or an external agent is required to do the work.

Be careful in using the above interpretation to interpret circuit voltage or p.d. I would use conventional current instead of electron current. Let says that the current is flowing from A to B, what this means is that the electric potential at B is lower than that at A or VB < VA.

When we use a voltmeter to measure the p.d. between A and B, we need to place the red coloured lead at point A and the black coloured lead at B in order for us to get a positive voltage reading. If we switch the positions for the red and black coloured lead, we would get a negative voltage reading.

Does it contradict with the above interpretation of positive and negative potential difference? Not really.
I have no idea what you're talking about - you lost me at ''Note that this electric force is conservative.'' N.B. I am not a physicist, I'm a biochemist. I took Physics at AS level and I know enough about electricity to set up and understand an agarose gel electrophoresis, but that's about it.
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