# Transfer of Electrical Charges

I'm having a lot of trouble understanding how charges transfer exactly. Suppose I have 3 particles $$a,b,c$$, $$a$$ is negatively charged, $$b$$ is positively charged and $$c$$ is neutral. We let the charges of $$a,b$$ be $$\vert a \vert > \vert b \vert$$. If say $$a$$ were to touch $$c$$ the neutrally charged particle and then $$c$$ were to touch $$b$$, what happens here?

$$a$$ and $$b$$ attract so I am assuming that $$c$$ picks up charge and will also attract $$b$$. But how much charge is transfer? Is it $$\frac{a}{2}$$ and would be something to be memorized? I'm curious why as well.

Thanks in advance for your help.

## 1 Answer

Suppose I have 3 particles a,b,c, a is negatively charged, b is positively charged and c is neutral. We let the charges of a,b be |a|>|b|. If say a were to touch c the neutrally charged particle and then c were to touch b, what happens here?

Elementary particles can't transfer charges by touching.

Let's consider composite particles (like atoms) that can transfer charges, and atom $$a$$ having greater negative charge than atom $$b$$ has positive. This is caused by atom $$a$$ having $$n$$ more electrons than protons, and atom $$b$$ having $$m$$ less electrons than protons, where $$n>m$$. If the pull from the nucleus of atom $$b$$ is strong enough, some electrons from atom $$a$$ will be attracted to atom $$b$$ strongly enough that they either leave atom $$a$$ completely, or they create a chemical bond with atom $$b$$ that will hold together by $$a$$'s electrons being attracted by $$b$$'s nucleus, and $$b$$'s electrons attracted by $$a$$'s nucleus.

In case of three atoms, $$a$$ might transfer some electrons to $$c$$, which then might transfer some to $$b$$.

In case of macroscopic objects, the charge is caused by charge carriers, such as free electrons (usually), ions (like in electrolytes or plasma) or holes (like in semiconductors).

For example, let's have two pieces of metal $$a$$ and $$b$$, $$a$$ with abundance of free electrons on the surface giving it negative charge, and $$b$$ with lack of free electrons on the surface giving it positive charge. When $$a$$ and $$b$$ are brought into physical contact, some free electrons from $$a$$'s surface move to $$b$$'s surface to equalize the electric potential of both $$a$$ and $$b$$.

But how much charge is transfer?

For simplicity, let's consider two spherical conductors $$a$$ and $$b$$. $$a$$ has charge $$-Q_a$$ and $$b$$ charge $$Q_b$$, where $$Q_a>Q_b$$. It can be shown that the electric potential of $$a$$ is $$V_a=k\frac{-Q_a}{r_a}$$ and the electric potential of $$b$$ is $$V_b=k\frac{Q_b}{r_b}$$, where $$r_a$$ is the radius of $$a$$, $$r_b$$ is the radius of $$b$$ and $$k$$ is Coulomb's constant. If we denote the charge moved from $$a$$ to $$b$$ by $$Q_\Delta$$, we get

$$k\frac{-Q_a-Q_\Delta}{r_a}=k\frac{Q_b+Q_\Delta}{r_b}$$

because the new potentials must be equal. By solving for $$Q_\Delta$$, we get

$$Q_\Delta = -\frac{Q_ar_b + Q_br_a}{r_a + r_b}$$

which is the charge that will move from $$a$$ to $$b$$.

• I'm looking up a homework solution and it starts out saying that the charge is split equally between $a$ and $c$ when they touch. Is this conceptual incorrect? here is a link [link]slader.com/textbook/… – dls Jan 31 at 22:47
• In the link, they're taking about spheres. Look at the equation I wrote for $Q_\Delta$: If you substitute $r_a = r_b$ and $Q_b = 0$, you'll get $Q_\Delta = -\frac{Q_a}2$, so for an uncharged $b$, $a$ gives away half of its charge, so their solution is correct. – Golden Gleam Feb 1 at 8:34
• (Btw, your link is broken and the correct link is here: slader.com/textbook/…) 🙂 – Golden Gleam Feb 1 at 8:58
• @dls Also, I'm pinging you now because I didn't realize before that I had to ping you for you to be notified of my comments, so just to make sure you'll notice them. – Golden Gleam Feb 1 at 11:57