Electric field inside a conductor

• In electrostatic equilibrium, the net electric field inside a conductor is zero.
• But the electric field due to individual charges is not zero.

Charge inside a conductor

• Consider a conductor of radius R which is given a charge \(+Q\), as shown in figure.

• If the conductor is in electrostatic equilibrium, then all the excess charges reside on the exterior surface of the conductor.

Proof

Since, Electric field inside conductor is zero, i.e., \(\vec E=0\)

Thus, \(\int\vec E\cdot d\vec A=0\) ....(1)

By Gauss's Law,

\(\int\vec E\cdot d\vec A=\dfrac {q_{inside}}{\epsilon_0}\)

\(0=\dfrac {q_{inside}}{\epsilon_0}\) [ From equation (1) ]

\(q_{inside}=0\)

Electric field at a point P inside a conductor due to a charge \(q\)

• Consider a point P inside the conductor. A charge \(+q\) is placed outside it, as shown in figure.

• Due to presence of charge \(+q\), the charges inside the conductor reside on the surface of conductor, as shown in figure.

• Thus, there is no charge at point P and hence no electric field.

• So, electric field due to \(+q\) at point P is \(\vec E_P=0\) .

• Consider a neutral conductor with a cavity inside it.
• A point charge \(+Q\) is placed inside the cavity, as shown in figure.

• Consider a Gaussian surface which completely encloses the cavity, as shown in figure.

• As electric field \(\vec E\) is zero in conductor, the value of \(\oint\vec E \cdot d\vec A\) on Gaussian surface will be zero.

\(\vec E=0\)

\(\Rightarrow \oint\vec E \cdot d\vec A=0\)

By Gauss's Law,

\(\oint\vec E \cdot d\vec A=\dfrac {q_{inside}}{\mathcal{\epsilon}_0}\)

\(0=\dfrac {q_{inside}}{\mathcal{\epsilon}_0}\)

\(q_{inside}=0 \)

• In order to make net charge inside the Gaussian surface zero, the equal negative charge gets distributed on the surface of cavity, as shown in figure.

• Since, the conductor is neutral. Thus to compensate the distribution of negative charge, equal positive charge gets distributed on the surface of conductor, as shown in figure.

Electric field inside a conductor

• In electrostatic equilibrium, the net electric field inside a conductor is zero.
• But the electric field due to individual charges is not zero.

Charge inside a conductor

• Consider a conductor of radius R which is given a charge \(+Q\), as shown in figure.

• If the conductor is in electrostatic equilibrium, then all the excess charges reside on the exterior surface of the conductor.

Proof

Since, Electric field inside conductor is zero, i.e., \(\vec E=0\).

Thus, \(\int\vec E\cdot d\vec A=0\) ....(1)

By Gauss's Law,

\(\int\vec E\cdot d\vec A=\dfrac {q_{inside}}{\epsilon_0}\)

\(0=\dfrac {q_{inside}}{\epsilon_0}\) [ From equation (1) ]

\(q_{inside}=0\)

Electric field at a point P inside a conductor due to a charge \(q\)

• Consider a point P inside the conductor. A charge \(+q\) is placed outside it, as shown in figure.

• Due to presence of charge \(+q\), the charges inside the conductor reside on the surface of conductor, as shown in figure.

Thus, there is no charge at point P and hence no electric field.

• So, electric field due to \(+q\) at point P is \(\vec E_P=0\).

• A part of surface of a charged conductor is shown in the figure.

• The electric field is always perpendicular to the surface, whether the conductor is positively or negatively charged.

• Now, Suppose \(\vec E_{outside}\) has a component tangent to the surface of charged conductor.
• A force on the surface charges would be exerted by this component.
• This force would cause surface charges to move, violating the electrostatic equilibrium.
• The only exterior electric field consistent with electrostatic equilibrium is one that is perpendicular to the surface of charged conductor.

• Consider a conductor having some positive charge, as shown in figure.
• Consider a cylindrical Gaussian Surface of cross-sectional area A, as shown in figure.
• The direction of electric field is perpendicular to the surface outside and \(\vec E_{inside}=0\).
• Total flux linked with the Gaussian Surface ,

•  \(Flux=E.A\;cos\,0^\circ\)

          \(=\,EA\)

  By Gauss's Law,

  \(\text{Flux}=\dfrac {\sum Q_{enclosed}}{\epsilon_0}\) ...(1)

• Charge enclosed by cylindrical Gaussian surface = \(\sigma A\)

• where, \(\sigma\) is surface charge density of the Gaussian surface where electric field is to be calculated

  Thus, \(\text {Flux}=\dfrac {\sigma A}{\epsilon_0}\) ...(2)

  Comparing equation (1) and (2)

  \(EA=\dfrac {\sigma A}{\epsilon_0}\)

  \(E=\dfrac {\sigma }{\epsilon_0}\)

  \(\vec E=\dfrac {\sigma}{\epsilon_0}\;(\hat n)\), where \(\hat n\) is the unit vector in the direction of electric field