Electric potential halfway between two equal but opposite charges

[LightningB0LT]

I'm having a hard time understanding why the electric potential halfway between two equal but opposite charges is 0. If I put a positive charge halfway between a 5 mC charge and a -5 mC charge, it will accelerate toward the negative charge, but how can it do this if it starts with no potential energy?

 

[jtbell]

how can it do this if it starts with no potential energy?

Consider an analogy with gravity instead of electric force. Let the gravitational potential energy of an object be zero at the Earth's surface (the usual U = mgh formula). Dig a hole in the ground. Hold an object over the hole, exactly in line with the Earth's surface around the hole. What is its potential energy there? Now, let it go. What happens to it?

 

[nasu]

It has potential energy. As it approaches the negative charge its potential energy decreases, becomes more negative. So it has more energy in the middle than when it gets closer.
The fact that it is zero is just a matter of choice. The variation is what matters.
You can make it 10 J if you want. Just add a constant to the potential energy.

 

[BiGyElLoWhAt]

The fact that it is zero is just a matter of choice.

What's the first thing you do in a potential situation?

*SPOILER*

Set your zero.
Whether its $U_{g}$ or $U_{q}$, you need to set your zero, if you call the hot plate +10V instead of +5, that makes your low V plate at 0, or you could call the high V plate 0 and the other one -10V. Tomatoe Tomahto.

 

[LightningB0LT]

Thanks, that makes it a lot easier to understand.

 

[BiGyElLoWhAt]

Glad to help :)

 

[ZapperZ]

nasu has stated this, but I think it needs to be reemphasized here. The "force" or the electric field is the gradient of the potential. In other words, it is how much the potential changes with position. So it isn't the absolute value at every field point that is important, but rather, how the value changes as the position changes.

You could be in a situation where you have a potential of 100V, but if it is 100V all over the place, the charge will still not experience a force. The gradient will be zero there and thus, no net electric field/force.

Zz.

 

[Zahid Iftikhar]

I want to comment on ZapperZ's last lines, i.e. the field will be zero where change in potential is zero. True this is what the electric intensity and potential gradient relation says. But for the current question, the electric field is nowhere zero between the charges. The test charge midway will have the tendency to move towards negative charge. So if E≠0 at the midpoint, then ΔV≠0, Still a lot of confusion for me. Please provide more help.

 

[ZapperZ]

I want to comment on ZapperZ's last lines, i.e. the field will be zero where change in potential is zero. True this is what the electric intensity and potential gradient relation says. But for the current question, the electric field is nowhere zero between the charges. The test charge midway will have the tendency to move towards negative charge. So if E≠0 at the midpoint, then ΔV≠0, Still a lot of confusion for me. Please provide more help.

1. You are replying to a thread that was last active in 2014.

2. I was emphasizing to the OP the general concept that the electric field has nothing to do with the value of the potential, but rather the GRADIENT of the potential. I bought up a counter example where the potential is not zero, but a constant everywhere. Here, even if the potential isn't zero, you can still get a zero electric field. It wasn't an explanation for the situation in the problem.

3. I never stated that the E-field is zero at the midpoint. In fact, the explanation that it is the gradient of the potential rather than the potential itself that is relevant assures that E-field is NOT zero at the midpoint.

Zz.

 

[Zahid Iftikhar]

Thank u ZapperZ for the kind reply.
Actually I consult PhysicsForum when any difficulty in explaining a physical concept arises and search old posts and mostly find stuff sufficient to my requirement. Sometime if I need more help and the thread is open I add my question. That is why I posted my own query in this thread.
This question has real confusion, at least for me. But I will not share my view unless I know the policy of the PF whether to start a new thread of my own with this question. Please guide me on this.
Regards
Zahid

 

[davenn]

But I will not share my view unless I know the policy of the PF whether to start a new thread of my own with this question. Please guide me on this

It is always preferred and wise to start your own thread. Reviving dead/old posts is definitely not encouraged.
In fact many of the older threads get locked to stop that occurring

Regards
Dave

 

[Zahid Iftikhar]

Sure I will... Thank you.