Skip to article frontmatterSkip to article content

01 Coulomb’s Law (2)

Aim

To show the inverse square distance dependence of the electrostatic force between point charges.

Subjects

Diagram

.

Figure 1:.

Equipment

Presentation

The balls are hanging and touch each other. The shadow of the ping-pong balls is projected on the blackboard.

By means of the Van de Graaff generator the two balls are charged and immediately they separate by electrostatic repulsion. While the shadows of the two balls are dancing around towards their equilibrium, the method of the demonstration is explained to the students and to them it is shown that when we suppose the power in Coulomb’s law (1785) is really -2 , the determined distance-ratio should be 21/3=1.2592^{1 / 3}=1.259 (see Explanation). When the two balls have come to rest, the centres of the shadows of the balls are chalk-marked on the blackboard.

Now the two threads are sandwiched at the halfway point by means of a sliding piece of tape. (This tape is fixed to the threads already before you start the experiment; see Figure 2.)

.

Figure 2:.

Clearly can be seen that the two shadows are closer to each other now. Again the centres of the ball-shadows are chalk-marked.

On the blackboard the two separations are measured. (In a trial, we measured 78and 62 cm62 \mathrm{~cm} respectively.) The ratio is calculated ( 78/62=1.25878 / 62=1.258 ). This value is very close to the value 1.259 mentioned before and so the power in Coulomb’s law being- 2 is supported by this demonstration.

This demonstration gives the opportunity to stress to the students that Coulomb’s law is empirical and in that way very fundamental to the theory of electromagnetism.

That’s why it is very fundamental that measurements around Coulomb’s law are still performed, trying to determine nn with increasing accuracy (nowadays -1971- it stands to be accurate to 1 part in 1016; between 1,99999999999999994 and 2,00000000000000058 )

Explanation

Figure 3 shows that in the equilibrium position: Fcoulomb =mgtanφF_{\text {coulomb }}=m g \tan \varphi.

.

Figure 3:.

Since tanφ=r/2y,FCoulomb =mgr2y\tan \varphi=r / 2 y, F_{\text {Coulomb }}=\frac{m g r}{2 y}

Supposing FCoulomb =kq2rnF_{\text {Coulomb }}=k \frac{q^{2}}{r^{n}}, our equilibrium equation is kq2rn=mgr2y\frac{k q^{2}}{r^{n}}=\frac{m g r}{2 y}.

k,q,mk, q, m, and gg are constants, so rn+1y\frac{r^{n+1}}{y} has a constant value (rn+1y=2ykq2mg\left(\frac{r^{n+1}}{y}=\frac{2 y k q^{2}}{m g}\right. )

The first measurement gives us r1n+1y1\frac{r_{1}^{n+1}}{y_{1}}, the second r2n+1y2\frac{r_{2}^{n+1}}{y_{2}}. So, r1n+1y1=r2n+1y2\frac{r_{1}^{n+1}}{y_{1}}=\frac{r_{2}^{n+1}}{y_{2}}.

Since in our demonstration y1=2y2y_{1}=2 y_{2}, we find r1n+1r2n+1=y1y2=2\frac{r_{1}^{n+1}}{r_{2}^{n+1}}=\frac{y_{1}}{y_{2}}=2, and so: r1r2=21n+1\frac{r_{1}}{r_{2}}=2^{\frac{1}{n+1}}

For different values of nn we calculate for r1r2\frac{r_{1}}{r_{2}} :

n\boldsymbol{n}r1r2\frac{r_{1}}{r_{2}}
...\ldots \ldots \ldots ..
11,414.1,414 ..
1.51.319.1.319 ..
21,259.1,259 ..
2.51.219.1.219 ..
31.189.1.189 ..
...\ldots \ldots \ldots ..

In this way measuring r1r_{1} and r2r_{2} will give us the value for nn.

The mentioned measurement in the Presentation with r1=78 cmr_{1}=78 \mathrm{~cm} and r2=62 cmr_{2}=62 \mathrm{~cm} presents the ratio r1r2=1,258\frac{r_{1}}{r_{2}}=1,258.. the table above shows that this produces a value for nn very close to 2 .

Remarks

Sources

Supplement

Historical results

E(r)1r2+εE(r)-\frac{1}{r^{2+\varepsilon}}