Electric Charges and Fields

Using Coulomb’s law,

$F = k\dfrac{q_1q_2}{r^2}$

Given: $q_1=2\times10^{-7}\,\text{C}$, $q_2=3\times10^{-7}\,\text{C}$, $r=30\,\text{cm}=0.30\,\text{m}$, and $k=9\times10^9\,\text{N m}^2\text{/C}^2$.

$$F=9\times10^9\times\frac{(2\times10^{-7})(3\times10^{-7})}{(0.30)^2}$$

$$=(9\times10^9)\times\frac{6\times10^{-14}}{0.09}$$

$$=9\times10^9\times6.67\times10^{-13} =6.0\times10^{-3}\,\text{N}$$

So the force is repulsive because both charges are positive.

By Coulomb’s law,

$$F=k\frac{|q_1q_2|}{r^2}$$

Given $F=0.2\,\text{N}$, $q_1=0.4\,\mu\text{C}=0.4\times10^{-6}\,\text{C}$, $q_2=0.8\,\mu\text{C}=0.8\times10^{-6}\,\text{C}$.

So,
$$r^2=k\frac{|q_1q_2|}{F} =9\times10^9\times\frac{(0.4\times10^{-6})(0.8\times10^{-6})}{0.2}$$

$$=9\times10^9\times\frac{0.32\times10^{-12}}{0.2} =9\times10^9\times1.6\times10^{-12} =1.44\times10^{-2}$$

$$r=\sqrt{1.44\times10^{-2}}=0.12\,\text{m}$$

So the distance is 0.12 m.

For the force on the second sphere, by Newton’s third law, it has the same magnitude and opposite direction. Since the charges are unlike, the force is attractive.

To check dimensions:

$$\frac{ke^2}{Gm_em_p}$$

- $k$ has units $\text{N m}^2\text{/C}^2$
- $e^2$ has units $\text{C}^2$
- $G$ has units $\text{N m}^2\text{/kg}^2$
- $m_em_p$ has units $\text{kg}^2$

So the units are

$$\frac{(\text{N m}^2\text{/C}^2)(\text{C}^2)}{(\text{N m}^2\text{/kg}^2)(\text{kg}^2)}=1$$

Hence the ratio is dimensionless.

From the chapter, the value is approximately

$$\frac{ke^2}{Gm_em_p}\approx 2.4\times10^{39}$$

This ratio signifies the relative strength of electric force to gravitational force between a proton and an electron. The electric force is about $10^{39}$ times stronger.

(a) Electric charge is quantised means the charge on any body is always an integral multiple of the elementary charge $e$:

$$q=ne, \quad n=0,\pm1,\pm2,\ldots$$

So charge does not vary continuously at the microscopic level; it exists in discrete units of $e$.

(b) For macroscopic bodies, the total charge contains an extremely large number of electrons or protons. Since $e=1.6\times10^{-19}\,\text{C}$ is very small, the step size is negligible compared with ordinary charges like microcoulombs. Therefore charge appears continuous and quantisation can be ignored at large scale.

This is consistent with conservation of charge because when two bodies are rubbed, no new charge is created or destroyed. Charges are only transferred from one body to the other. For example, when a glass rod is rubbed with silk, some electrons move from the glass to the silk. So the glass becomes positively charged and the silk negatively charged, but the total charge of the combined system remains the same.

At the centre of the square, the charges at opposite corners are at equal distances from the centre.

- The two +2 μC charges are at opposite corners, so their forces on the central test charge are equal and opposite.
- The two −5 μC charges are also at opposite corners, so their forces on the central test charge are equal and opposite.

Therefore all forces cancel pairwise, and the net force is zero.

(a) A field line is drawn so that its tangent at every point gives the direction of the electric field. Since the electric field at a point has a definite direction, the field line must be a continuous curve. A sudden break would mean the field has no direction at that point, which is not possible.

(b) Two field lines can never cross because if they did, the electric field at the point of intersection would have two directions at the same time, which is impossible. The field at any point can have only one unique direction.

The midpoint is $0.10\,\text{m}$ from each charge.

For the positive charge $+3\,\mu\text{C}$, the field at O is away from the positive charge, i.e. towards B.
For the negative charge $-3\,\mu\text{C}$, the field at O is towards the negative charge, i.e. also towards B.
So the fields add.

Magnitude due to one charge:
$$E_1=\frac{kq}{r^2}=\frac{9\times10^9\times3\times10^{-6}}{(0.10)^2} =2.7\times10^6\,\text{N/C}$$

Total field:
$$E=2E_1=5.4\times10^6\,\text{N/C}$$

directed from the positive charge to the negative charge.

For the test charge $q_0=-1.5\times10^{-9}\,\text{C}$,
$$F=q_0E$$
Magnitude:
$$|F|=(1.5\times10^{-9})(5.4\times10^6)=8.1\times10^{-3}\,\text{N}$$
Since the test charge is negative, the force is opposite to the field direction, i.e. towards the positive charge.

Total charge:
$$q_{\text{total}}=2.5\times10^{-7}+(-2.5\times10^{-7})=0$$

Dipole moment is
$$\mathbf{p}=q\times 2a\,\hat{p}$$
Here the charges are on the $z$-axis at $-15\,\text{cm}$ and $+15\,\text{cm}$, so separation is
$$2a=30\,\text{cm}=0.30\,\text{m}$$
Magnitude:
$$p=(2.5\times10^{-7})(0.30)=7.5\times10^{-8}\,\text{C m}$$
Direction is from negative charge to positive charge, i.e. along +z-axis.

Torque on a dipole in a uniform field is

$$\tau=pE\sin\theta$$

Given $p=4\times10^{-9}\,\text{C m}$, $E=5\times10^4\,\text{N/C}$, and $\theta=30^\circ$.

$$\tau=(4\times10^{-9})(5\times10^4)\sin30^\circ$$

$$=2\times10^{-4}\times\frac{1}{2} =1\times10^{-4}\,\text{N m}$$

So the torque is **$1\times10^{-4}$ N m**.

(a) Number of electrons transferred:

$$n=\frac{Q}{e}=\frac{3\times10^{-7}}{1.6\times10^{-19}}$$

$$=1.875\times10^{12}$$

Since the polythene becomes negatively charged, it must have gained electrons from the wool.

(b) Yes, there is a transfer of mass because electrons have mass. But the transferred mass is extremely small, so in practice it is negligible.

(a) Using Coulomb’s law,

$$F=k\frac{q^2}{r^2}$$

Given $q=6.5\times10^{-7}\,\text{C}$ and $r=50\,\text{cm}=0.50\,\text{m}$.

$$F=9\times10^9\times\frac{(6.5\times10^{-7})^2}{(0.50)^2}$$

$$=(9\times10^9)\times\frac{4.225\times10^{-13}}{0.25} =1.521\times10^{-2}\,\text{N}$$

So, approximately
$$F\approx 1.5\times10^{-2}\,\text{N}$$

(b) If each charge is doubled, force becomes 4 times.
If distance is halved, force becomes 4 times more.
So total increase = $4\times4=16$ times.

$$F' = 16F \approx 16(1.5\times10^{-2}) = 2.4\times10^{-1}\,\text{N}$$

So the new force is about **$2.4\times10^{-1}$ N**.