Electrostatics of Conductors (Conductors In Electrostatic Field)}
Discuss the behaviour of an conductor in an electrostatic field. Derive expression for electric field at the surface of a charged conductor.
Behaviour of Conductors in the Electrostatic Field
1. Net electrostatic field inside a conductor is zero.
A conductor contains large number of free electrons. When it is placed in the electrostatic field ($\vec{E}_0$), each electron experiences a force ($\vec{F} = -e\vec{E}_0$) in a direction opposite to the direction of applied field $\vec{E}_0$. This force causes free electrons in the conductor to move in a direction opposite to the direction of the applied electric field $\vec{E}_0$. Therefore, side A of the conductor becomes positively charged due to the transfer of free electrons to the side B and the side B of the conductor becomes negatively charged. In other words, side A of the conductor has less electrons and side B of the conductor has more electrons.
This re-distribution of charges creates its own additional electric field called induced electric field ($\vec{E}_p$). The induced electric field acts in a direction opposite to the direction of external field $\vec{E}_0$ but its magnitude is equal to the magnitude of the external electric field inside the conductor. Hence, net electric field, $\vec{E}_{net}$ in the interior of the conductor placed in the electrostatic field is zero. That is, $\vec{E}_{net} = \vec{E}_0 + \vec{E}_p = 0$
2. Electrostatic field just outside the charged conductor is perpendicular to the surface of the conductor.
Suppose, electrostatic field ($\vec{E}$) is not perpendicular to the surface of a conductor. Let $E$ make an angle $\theta$ with the surface of the conductor, then $E \cos \theta$ is tangential component of the field along the surface of the conductor.
This tangential component of electrostatic field will cause flow of charges on the surface of the conductor leading to the surface current. But there is no surface current in electrostatics because conductor placed in the electrostatic field has only static charges. Thus, electric field just outside the surface can only have a normal component and no tangential component.
That is, $E \cos \theta = 0$ or $\cos \theta = 0$ or $\theta = 90^\circ$ because $E \neq 0$
which clearly shows that electrostatic field just outside a charged conductor is perpendicular to the surface of the conductor at every point.
The electric field lines start and end perpendicular to the surface of a conductor.
3. Net charge in the interior of a charged conductor is zero.
According to Gauss' theorem,
$\oint_S \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0}$
Since electric field $\vec{E} = 0$ inside a conductor, so the net electric charge inside a conductor is zero.
That is,
$q = 0$
4. Charge resides on the surface of a conductor.
Consider a conductor suspended with an insulating thread. Let this conductor be given an excess charge $q$. Imagine a Gaussian surface which lies just inside and near to the surface of the conductor.
Since electric field ($\vec{E}$) in the interior of the conductor is zero, so it must also be zero for all points on the Gaussian surface which is also inside the conductor.
According to Gauss' law, the net charge inside the Gaussian surface must be zero. If excess charge $q$ is not inside the Gaussian surface, then it must lie on the surface of the conductor. Thus, charge can not reside in the interior of a conductor, it can only reside on the surface of the conductor.
5. Electrostatic potential is constant throughout the volume of a charged conductor and has the same value on its surface as inside it.
Electrostatic field inside a charged conductor is zero ($\vec{E} = 0$).
Therefore, $\frac{dV}{dr} = -E = 0$.
Hence, electrostatic potential
potential (V) is constant inside the conductor. In other words, the inner surface of the conductor is equipotential.
The electrostatic field on the surface of the conductor is normal to the surface of the conductor. Therefore, no work is done to move a test charge on the surface of conductor. In other words, $W = \Delta V q_0 = 0$. This is possible if every point on the surface of the conductor has same potential. Thus, surface of the conductor is equipotential surface. Therefore, we conclude that electrostatic potential is constant throughout the volume of a charged conductor and has the same value on its surface as inside it.
Every conductor is an equipotential volume (three-dimensional) rather than just an equipotential surface (two-dimensional).
6. Surface charge distribution may be different at different points on the conductor.
Consider a conductor of the shape shown in figure 31. Surface charge density ($\sigma$) is given by $\sigma = \frac{q}{A}$, where A is surface area.
The surface area of the conical or pointed end of the conductor is smaller than that of the less curved part of the conductor. Hence, the surface charge density of the conical or pointed part of the conductor will be higher. Thus, the charge on the given conductor is not uniformly distributed. The charge at the pointed end is much more than the remaining surface.
7. Expression for the Electric field at the surface of charged conductor.
Consider a charged conductor of irregular shape. Let $\sigma$ be the surface charge density of the conductor. Consider a gaussian surface in the form of a cylinder having some portion lying inside the conductor.
The Gaussian cylinder has three parts, two circular ends and one curved portion. Since, electric field $\vec{E} = 0$ inside the conductor, so no electric flux is linked with one circular end which is inside the conductor. Also, no electric flux is linked with the curved portion of the cylinder as electric field $\vec{E}$ is perpendicular to this part.
Electric flux passes only through the circular end (of area vector $d\vec{S}$) of the Gaussian cylinder which is outside the conductor.
According to Gauss' law,
$\oint \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0}$
or
$\oint EdS \cos 0^\circ = \frac{q}{\epsilon_0} \quad \text{or} \quad \oint EdS = \frac{q}{\epsilon_0}$
Now
$\oint dS = dS \quad \text{and} \quad q = \sigma dS$
$\therefore \quad EdS = \frac{\sigma dS}{\epsilon_0} \quad \text{or} \quad E = \frac{\sigma}{\epsilon_0}$
Since electric field is perpendicular to the surface of the conductor, therefore, $\vec{E} = \frac{\sigma}{\epsilon_0} \hat{n}$, where $\hat{n}$ is the unit vector normal to the surface of the conductor in the outward direction.