Electric Charges

Gauss’s law

The total electric flux through any closed surface is equal to (\dfrac{1}{\varepsilon_0}) times the net electric charge enclosed by the surface.

[ \oint \vec{E}\cdot d\vec{A}=\frac{q_{\text{enclosed}}}{\varepsilon_0} ]

where (\varepsilon_0) is the permittivity of free space.

Electric field due to an infinitely long uniformly charged straight wire

Let an infinitely long straight wire have uniform linear charge density (\lambda). We have to find the electric field at a perpendicular distance (r) from the wire.

Consider a cylindrical Gaussian surface of radius (r) and length (l), coaxial with the wire.

By symmetry, the electric field is directed radially outward at every point on the curved surface of the cylinder and has the same magnitude (E).

1. Flux through the curved surface

[ \phi_{\text{curved}}=E(2\pi r l) ]

2. Flux through the two flat ends

The electric field is radial, while the area vectors of the flat ends are along the length of the wire. Hence,

[ \phi_{\text{ends}}=0 ]

So total flux,

[ \phi=E(2\pi r l) ]

3. Charge enclosed by the Gaussian surface

[ q_{\text{enclosed}}=\lambda l ]

4. Applying Gauss’s law

[ E(2\pi r l)=\frac{\lambda l}{\varepsilon_0} ]

[ E=\frac{\lambda}{2\pi\varepsilon_0 r} ]

In vector form, if (\hat{r}) is the unit vector perpendicular to the wire,

[ \vec{E}=\frac{\lambda}{2\pi\varepsilon_0 r}\hat{r} ]

The field is radially outward for positive (\lambda) and radially inward for negative (\lambda).

Numerical calculation

Given:

[ \lambda=2\times 10^{-6}\ \text{C m}^{-1} ]

[ r=10\ \text{cm}=0.1\ \text{m} ]

[ \frac{1}{4\pi\varepsilon_0}=9\times 10^9\ \text{N m}^2\text{C}^{-2} ]

We know,

[ E=\frac{\lambda}{2\pi\varepsilon_0 r} ]

Since (\dfrac{1}{2\pi\varepsilon_0}=2\left(\dfrac{1}{4\pi\varepsilon_0}\right)),

[ E=\frac{2\lambda}{4\pi\varepsilon_0 r} ]

[ E=\frac{2\times 9\times 10^9\times 2\times 10^{-6}}{0.1} ]

[ E=\frac{36\times 10^3}{0.1} ]

[ E=3.6\times 10^5\ \text{N C}^{-1} ]

[ \boxed{E=3.6\times 10^5\ \text{N C}^{-1}} ]

The electric field is directed radially outward from the wire because the charge is positive.

The charged comb induces opposite charge on the near side of each neutral paper bit and the same charge on the far side. This happens because the charges in the paper shift slightly (polarization).

The field of the comb is non-uniform, being stronger nearer to it. The nearer, oppositely charged side is attracted more strongly than the far side is repelled, so there is a net force of attraction and the paper bit is pulled to the comb even though its total charge is zero.

(d) $0.624\times10^{19}$ (that is, about $6.24\times10^{18}$ electrons)

$n=\frac{Q}{e}=\frac{1}{1.6\times10^{-19}}\approx6.25\times10^{18}$

This is the value in option (d), whose power of ten appears to be misprinted: $0.624\times10^{19}=6.24\times10^{18}$. No other option is close.

(a) Charge is not a base quantity in SI. The base quantity is current (ampere), and charge is obtained from it by $Q=It$ (coulomb = ampere second), so charge is derived.

(b) Take $e=1.6\times10^{-19}\ \text{C}$, $m_e=9.11\times10^{-31}\ \text{kg}$, $k=9.0\times10^{9}\ \text{N m}^2\text{C}^{-2}$, $G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$.

$\frac{F_e}{F_g}=\frac{ke^2/x^2}{Gm_e^2/x^2}=\frac{ke^2}{Gm_e^2}=\frac{9.0\times10^{9}\,(1.6\times10^{-19})^2}{6.67\times10^{-11}\,(9.11\times10^{-31})^2}\approx 4\times10^{42}$

Both forces vary as $1/x^2$, so the ratio is independent of $x$. The electric force is about $4\times10^{42}$ times the gravitational force, and (for two electrons) it is repulsive while gravity is attractive.

(c) Gravitational force between charged particles is smaller than the electric force by a factor of about $10^{42}$, so it is completely negligible and can be ignored in such calculations.

(d) ( 5.0 \times 10^{-19} , \text{C} )

By quantisation of charge, ( q = ne ), so the charge must be an integral multiple of ( e = 1.6 \times 10^{-19} , \text{C} ). Here ( 1.6 ), ( 3.2 ) and ( 4.8 \times 10^{-19} , \text{C} ) equal ( 1e ), ( 2e ) and ( 3e ) respectively, but ( \dfrac{5.0 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.125 ), which is not an integer — so this charge cannot exist on an isolated body.

Coulomb’s law

The electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges.

[ F \propto q_1q_2,\qquad F \propto \frac{1}{r^2} ]

[ \boxed{F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}} ]

where (\frac{1}{4\pi\varepsilon_0}=9\times 10^9\ \text{N m}^2\text{C}^{-2}) in vacuum.

Vector form

Let (\vec r_1) and (\vec r_2) be the position vectors of charges (q_1) and (q_2), and let

[ \vec r_{12}=\vec r_2-\vec r_1,\qquad r=|\vec r_{12}| ]

Then the force on (q_2) due to (q_1) is

[ \boxed{\vec F_{21}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^3}(\vec r_2-\vec r_1)} ]

and the force on (q_1) due to (q_2) is

[ \boxed{\vec F_{12}=-\vec F_{21}} ]

Numerical

Given:

[ q_1=+4\ \mu\text{C}=4\times 10^{-6}\ \text{C} ]

[ q_2=-3\ \mu\text{C}=-3\times 10^{-6}\ \text{C} ]

[ r=20\ \text{cm}=0.20\ \text{m} ]

[ \frac{1}{4\pi\varepsilon_0}=9\times 10^9\ \text{N m}^2\text{C}^{-2} ]

Force between (q_1) and (q_2):

[ F=\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^2} ]

[ F=\frac{9\times 10^9\times 4\times 10^{-6}\times 3\times 10^{-6}}{(0.20)^2} ]

[ F=\frac{0.108}{0.04}=2.7\ \text{N} ]

Since the charges are of opposite signs, the force is attractive.

[ \boxed{F=2.7\ \text{N}\ \text{(attractive)}} ]

Now a third charge

[ q_3=+2\ \mu\text{C}=2\times 10^{-6}\ \text{C} ]

is placed at the midpoint. Therefore,

[ r_{13}=r_{23}=0.10\ \text{m} ]

Force on (q_3) due to (q_1):

[ F_1=\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_3|}{r_{13}^2} ]

[ F_1=\frac{9\times 10^9\times 4\times 10^{-6}\times 2\times 10^{-6}}{(0.10)^2} ]

[ F_1=7.2\ \text{N} ]

Since (q_1) and (q_3) are both positive, this force is repulsive, i.e. directed away from (q_1) towards (q_2).

Force on (q_3) due to (q_2):

[ F_2=\frac{1}{4\pi\varepsilon_0}\frac{|q_2q_3|}{r_{23}^2} ]

[ F_2=\frac{9\times 10^9\times 3\times 10^{-6}\times 2\times 10^{-6}}{(0.10)^2} ]

[ F_2=5.4\ \text{N} ]

Since (q_2) is negative and (q_3) is positive, this force is attractive, i.e. directed towards (q_2).

Both forces on (q_3) act in the same direction, from (q_1) towards (q_2). Hence,

[ F_{\text{net}}=F_1+F_2=7.2+5.4=12.6\ \text{N} ]

[ \boxed{F_{\text{net}}=12.6\ \text{N}\ \text{towards }q_2} ]

So the net force on the midpoint charge is 12.6 N directed from (q_1) to (q_2).

Electric dipole moment: An electric dipole is a pair of equal and opposite point charges, (+q) and (-q), separated by a small distance. The electric dipole moment is defined as the product of the magnitude of either charge and the distance between them, directed from the negative charge to the positive charge.

$\vec{p} = q(2\vec{a})$

where (2a) is the dipole length. Its SI unit is the coulomb metre ((\text{C},\text{m})).

Electric field on the axial line

Let an electric dipole of charges (+q) and (-q) separated by distance (2a) have its centre at (O). Let (P) be a point on the axial line at distance (r) from (O), on the side of (+q).

Distance of (P) from (+q) is ((r - a)) and from (-q) is ((r + a)).

Field at (P) due to (+q) (away from (+q)):

$E_1 = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r-a)^2} \quad \text{(along } OP\text{)}$

Field at (P) due to (-q) (towards (-q)):

$E_2 = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r+a)^2} \quad \text{(along } PO\text{)}$

Since (E_1 > E_2), the resultant field is along (OP):

$E = E_1 - E_2 = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}\right]$

$E = \frac{q}{4\pi\varepsilon_0}\cdot\frac{(r+a)^2 - (r-a)^2}{(r^2-a^2)^2} = \frac{q}{4\pi\varepsilon_0}\cdot\frac{4ar}{(r^2-a^2)^2}$

Since (p = q(2a)), i.e. (q = \dfrac{p}{2a}):

$E = \frac{1}{4\pi\varepsilon_0}\cdot\frac{(p/2a)\cdot 4ar}{(r^2-a^2)^2}$

$\boxed{E = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2pr}{(r^2-a^2)^2}}$

The direction of (E) is along the dipole moment (\vec{p}).

For (r \gg a): Neglecting (a^2) in comparison with (r^2),

$E = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2pr}{r^4}$

$\boxed{E = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2p}{r^3}}$

Thus the axial field of a short dipole varies inversely as the cube of the distance from its centre.

Friction between the vehicle (and the flowing fuel) and the air produces static charge, which builds up on the body. A discharge with a spark could ignite the flammable material. The metallic rope touching the ground gives the charge a conducting path to the earth, so it leaks away continuously and no dangerous potential builds up.