# Why positive or negtive pole of battery can not attract small objects ?

magnetar
why positive or negtive polar of battery can not attract small objects ?

Gold Member
you tell me why do you think it should?

Homework Helper
Electrical and magnetic fields are not the same thing.

Gold Member
the terminals of a battery are at different potentials, that doesn't mean they would attract objects, they are not charged.

magnetar
you tell me why do you think it should?

Because the positive pole accumulate positive charges, it should attract small objects as the charged rubber rod do ?

Gold Member
no, thats what i said...they don't accumulate positive charges. they are not charged. just that positive terminal is at higher potential than negative one. no charges.

Homework Helper
i think what magnetar is asking is, if the charged ions or whatever they are are inside the battery are attracted to an electrode, why isn't anything outside the battery similarly attracted? Homework Helper
I didn't realise you meant static related attraction. It's because a normal battery doesn't have enough voltage to attract small objects like a statically charged rod with high voltage.

Homework Helper
It's because a normal battery doesn't have enough voltage to attract small objects like a statically charged rod with high voltage.

Now I'm confused. Voltage is relative.

One electrode has a potential difference of (say) 1.5 V from the other electrode.

But it has a high potential difference from "a statically charged rod with high voltage" …

so shouldn't that make it easier to attract an electron? If we bring an insulator with one free electron on its surface toward a battery, why isn't that attracted to one of the electrodes?

torquil
A battery is similar to an electric dipole. The potential difference between the battery poles is accompanied by an internal electric field. The interal electric field is caused by the electric charge polarization within the battery. Just like a capacitor.

A battery will have a nonzero surrounding electric field just like a capacitor, and will therefore (in principle) attract dielectric substances such as dust on both sides.

This was my understanding at least.

Homework Helper
I didn't realise you meant static related attraction. It's because a normal battery doesn't have enough voltage to attract small objects like a statically charged rod with high voltage.

Voltage is relative.
Assuming a moderate amount of capacitance, then a high voltage battery (if such a thing exists) would have signicant charge on at least at one of it's terminals. As mentioned above, batteries do attract dust.

phyeinstein_c
i think its bcoz of the chemical inside... that even though it has static electricity in it...it has electric field not magnetic... if attached to a solenoid then it attracts objects like paper

Gold Member
Most batteries are electrochemical cells, with voltage controlled by varying the concentration of the ions. the EMF can be calculated using the Nernst equation.

when a conductor connects the two terminals, the electrons get a path to go to a higher potential (which they die to do). the battery creates an electrostatic field outside it, which makes the electrons move.

phyeinstein_c
Most batteries are electrochemical cells, with voltage controlled by varying the concentration of the ions. the EMF can be calculated using the Nernst equation.

when a conductor connects the two terminals, the electrons get a path to go to a higher potential (which they die to do). the battery creates an electrostatic field outside it, which makes the electrons move.
and this causes magnetic field? Gold Member
NO. no magnetic field is created by the battery. but moving charges through the conductor creates a magnetic field.

phyeinstein_c
hooww can battery create mg field?????????? i stil dont get it

torquil
A battery not in a closed circuit (i.e. without a current running through it) does not create a magnetic field. I don't think anyone claimed that it did?

Homework Helper
A battery not in a closed circuit (i.e. without a current running through it) does not create a magnetic field. I don't think anyone claimed that it did?
No one did. What was stated the terminals of a battery have a static charge, relative to the voltage (divided by the equivalent of capacitance) of the battery, and that the static charge is enough to attract dust.

torquil
No one did. What was stated the terminals of a battery have a static charge, relative to the voltage (divided by the equivalent of capacitance) of the battery, and that the static charge is enough to attract dust.

Yeah, I stated that myself in post #10. I wrote post #17 since I didn't understand the reason for the confusion in post #16...

Gold Member
i said it doesn't create any magnetic field.

Gold Member
Because the positive pole accumulate positive charges, it should attract small objects as the charged rubber rod do ?

You're right, in principle the battery should attract or repel charged objects. The problem is that the amount of charge will be extremely tiny for a typical battery.

If you look at a normal battery, it has metal plates at the top and bottom. This essentially makes the battery a capacitor. The capacitance of a parallel plate capacitor is
$$C=\epsilon_{0}\epsilon_{r}\frac{A}{d}=\frac{Q}{V}$$
(This formula neglects edge effects which will be very prominent in a typical, say AA battery, but I just want to get an order of magnitude estimate)
In a device designed to be a capacitor, the distance between the plates will usually be extremely small, and filled with a material with a high dielectric constant to make Q relatively large. In a battery, 'd' will be something like 5cm, and the dielectric constant may or may not be significant, I'm not sure. For water the dielectric constant will be something like 80, which is relatively high, and you will see that this will not really make a difference in the fact that the force is extremely small. Let's take A=1cm^2, dielectric constant = 1, d = 5cm, V=1.5V. We get
$$Q \approx 2.6\cdot 10^{-14}coulombs$$
The force between two batteries separated by 1cm will be:
$$F=\frac{Q^2}{4\pi\epsilon_{0}r^2}\approx 6.1\cdot 10^{-14}N$$
So you can see the force is microscopic. Changing the dielectric constant to 100 could bring this up to 6*10-10, which is much bigger but still not readily observable.

phyeinstein_c