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Conductors

Practice magnitude of electric field just outside the conductor, electric field due to the point charge inside the cavity. Learn how to calculate electric field due to a charged conductor.

Electric Field due to a Charged Conductor

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 \(\vec E_P=0\) .

Illustration Questions

A conductor is given a charge \(+Q\).  A point P is situated inside it, as shown in figure. The value of electric field at P is

A Zero

B \(\dfrac {1}{4\pi\epsilon_0}\,\cdot \dfrac {Q}{r^2}\)

C \(\dfrac {1}{4\pi\epsilon_0}\,\cdot \dfrac {Q}{R^2}\)

D None of these

×

The charge \(+Q\) resides on the surface of conductor thus, the net charge inside the conductor is zero.

image

The net electric field inside the conductor is zero.

Thus, at point P the electric field is zero.

image

A conductor is given a charge \(+Q\).  A point P is situated inside it, as shown in figure. The value of electric field at P is

image
A

Zero

.

B

\(\dfrac {1}{4\pi\epsilon_0}\,\cdot \dfrac {Q}{r^2}\)

C

\(\dfrac {1}{4\pi\epsilon_0}\,\cdot \dfrac {Q}{R^2}\)

D

None of these

Option A is Correct

Direction of Electric Field at the Surface of Charged Conductor

  • 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.

Illustration Questions

A uniform electric field is shown in the figure. A rectangular conductor is placed in this field. Choose the correct option regarding the field lines after the presence of the conductor.

A

B

C

D None of these

×

Since, the electric field is always perpendicular to the surface of the conductor and  \(\vec E_{inside}=0\).

Hence, option (A) is correct.

A uniform electric field is shown in the figure. A rectangular conductor is placed in this field. Choose the correct option regarding the field lines after the presence of the conductor.

image
A image
B image
C image
D

None of these

Option A is Correct

Magnitude of Electric Field just outside the 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

Illustration Questions

A conductor is shown in figure, which has charge \(Q=8.85×10^{-22}\,C\), distributed over the surface of area \(A=10^{-10}\,m^2\). The value of electric field at P will be (if P is just outside the surface) Given:  \(\epsilon_0=8.85×10^{-12}\,C^2/Nm^2\)

A \(20\,N/C\)

B \(30\,N/C\)

C \(1\,N/C\)

D \(Zero\)

×

Electric field just outside the conductor,

\(\vec E_P=\dfrac {\sigma}{\epsilon_0}\)

where  \(\sigma\)= surface charge density

\(\epsilon_0\)= permittivity of free space

Electric field at point P

\(\vec E_P=\dfrac {Q}{A\,\epsilon_0}\) \(\left [ \because \sigma =\dfrac {Q}{A}\right]\)

image

Given:  \(Q=8.85×10^{-22}\,C\)\(A=10^{-10}\,m^2\)

\(\epsilon_0 =8.85×10^{-12}\,C^2/Nm^2\)

\(\vec E_P=\dfrac{8.85\times 10^{-22}}{8.85\times10^{-12}\times10^{-10} }\)

\(\vec E_P=1\,N/C\)

image

A conductor is shown in figure, which has charge \(Q=8.85×10^{-22}\,C\), distributed over the surface of area \(A=10^{-10}\,m^2\). The value of electric field at P will be (if P is just outside the surface) Given:  \(\epsilon_0=8.85×10^{-12}\,C^2/Nm^2\)

image
A

\(20\,N/C\)

.

B

\(30\,N/C\)

C

\(1\,N/C\)

D

\(Zero\)

Option C is Correct

Electric Field due to the Point Charge inside the Cavity

  • 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.

Illustration Questions

A neutral conductor with a cavity inside it, is shown in figure. A charge \(-Q\) is placed inside the cavity. Which one of the following is the correct arrangement of charge distribution?

A

B

C

D None of these

×

As electric field inside a conductor is always zero, thus the net charge inside the Gaussian surface has to be zero.

image

Gaussian surface should be selected such that point P resides on it.

Thus, the equal positive charge gets distributed on the surface of cavity.

image

Since, the conductor is neutral, thus the outer surface of the conductor acquires equal negative charge, as shown.

image

A neutral conductor with a cavity inside it, is shown in figure. A charge \(-Q\) is placed inside the cavity. Which one of the following is the correct arrangement of charge distribution?

image
A image
B image
C image
D

None of these

Option B is Correct

Distribution of Charge on the Surface of a Conductor

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 \(\vec E_P=0\).

Illustration Questions

A conductor of spherical shape is given a charge \(+Q\) and kept in isolation. Choose the correct option.

A Charge will distribute on its surface uniformly

B Charge will distribute throughout the volume of conductor uniformly

C Charge will distribute throughout the volume of conductor non-uniformly

D Charge will collect at a point on surface

×

In electrostatic equilibrium, the net charge inside a conductor is zero and all the excess charges reside on the surface of the conductor, uniformly.

Hence, option (A) is correct.

A conductor of spherical shape is given a charge \(+Q\) and kept in isolation. Choose the correct option.

A

Charge will distribute on its surface uniformly

.

B

Charge will distribute throughout the volume of conductor uniformly

C

Charge will distribute throughout the volume of conductor non-uniformly

D

Charge will collect at a point on surface

Option A is Correct

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