Question

In the figure below, a small circular hole of radius R = 1.80 cm has been cut in the middle of an infinite, flat, nonconducting surface that has uniform charge density σ = 6.70 pC/m². A z-axis, with its origin at the hole's center, is perpendicular to the surface. In unit-vector notation, what is the electric field at point P at z = 2.68 cm? (Hint: See Eq. 22-26 and use superposition.)
_________N/C

Answer 1

Concepts and reason

The concepts required to solve the given problem are, electric field around an infinite plane with uniform charge density and electric field for a uniformly charged disk along its axis.

For a given charge density, use the expression of electric field for an infinite plane and electric field for a disk at a distance z, to determine an expression for electric field vector at the desired point P. Next, for the given values of $\sigma ,z,{\rm{ and }}r$ evaluate the expression thus obtained to solve for the electric field vector at P.

Fundamentals

On an xy plane, electric field around an infinite flat surface with uniform charge density is expressed as,

${\vec E_z} = + \left( {2\pi {k_e}\sigma } \right)\hat k$

Here, ${\vec E_z}$ is the electric field vector along an axis normal to the surface of the plane, ${k_e}$ is the Coulomb constant, $\hat k$ is the unit vector along z axis, and $\sigma$ denotes surface charge density.

The above expression of electric field is valid only for $z > 0$.

On an xy plane, electric field for a uniformly charged disk along its axis is expressed as,

${\vec E_z} = 2\pi {k_e}\sigma \left( {1 - \frac{z}{{\sqrt {{z^2} + {r^2}} }}} \right)\hat k$

Here, z is the distance along positive z axis and r is the radius of the disk.

The electric field vector for an infinite flat surface with uniform charge density is,

${\vec E_{z,p}} = \left( {2\pi {k_e}\sigma } \right)\hat k$

Here, ${\vec E_{z,p}}$ is the electric field vector along an axis normal to the surface of the infinite plane.

The electric field vector for a given hole is expressed as,

${\vec E_{z,h}} = 2\pi {k_e}\left( { - \sigma } \right)\left( {1 - \frac{z}{{\sqrt {{z^2} + {r^2}} }}} \right)\hat k$

Here, the hole is considered to have charge density of $- \sigma$, ${\vec E_{z,h}}$ is the electric field vector for the hole, $r$is the radius of the hole, and z is the distance of point P along z axis, from the center of the hole.

Calculate the Net electric field for the given infinite flat surface with hole as,

${\vec E_z} = {\vec E_{z,p}} + {\vec E_{z,h}}$

Substitute, $\left( {2\pi {k_e}\sigma } \right)\hat k$ for ${\vec E_{z,p}}$ and $2\pi {k_e}\left( { - \sigma } \right)\left( {1 - \frac{z}{{\sqrt {{z^2} + {r^2}} }}} \right)\hat k$ for ${\vec E_{z,h}}$.

$\begin{array}{c}\\{{\vec E}_z} = \left( {2\pi {k_e}\sigma } \right)\hat k + 2\pi {k_e}\left( { - \sigma } \right)\left( {1 - \frac{z}{{\sqrt {{z^2} + {r^2}} }}} \right)\hat k\\\\ = \left( {2\pi {k_e}\sigma } \right)\hat k - 2\pi {k_e}\sigma \left( {1 - \frac{z}{{\sqrt {{z^2} + {r^2}} }}} \right)\hat k\\\\ = \left( {2\pi {k_e}\sigma } \right)\left( {1 - 1 + \frac{z}{{\sqrt {{z^2} + {r^2}} }}} \right)\hat k\\\end{array}$

Thus,

${\vec E_z} = \left( {2\pi {k_e}\sigma } \right)\left( {\frac{z}{{\sqrt {{z^2} + {r^2}} }}} \right)\hat k$ …… (1)

From equation (1),

${\vec E_z} = \left( {2\pi {k_e}\sigma } \right)\left( {\frac{z}{{\sqrt {{z^2} + {r^2}} }}} \right)\hat k$

Substitute, $8.99 \times {10^9}\frac{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ for ${k_e}$, $1.80{\rm{ cm}}$for $r$, $6.70\frac{{{\rm{pC}}}}{{{{\rm{m}}^{\rm{2}}}}}$ for $\sigma$, and $2.68{\rm{ cm}}$for $z$.

$\begin{array}{l}\\{{\vec E}_z} = \left( {2\pi \left( {8.99 \times {{10}^9}\frac{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}} \right)\left( {6.70\frac{{{\rm{pC}}}}{{{{\rm{m}}^{\rm{2}}}}} \times \frac{{{{10}^{ - 12}}{\rm{C}}}}{{1{\rm{ pC}}}}} \right)} \right)\left( {\frac{{\left( {2.68{\rm{ cm}} \times \frac{{{{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)}}{{\sqrt {{{\left( {2.68{\rm{ cm}} \times \frac{{{{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)}^2} + {{\left( {1.80{\rm{ cm}} \times \frac{{{{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)}^2}} }}} \right)\hat k\\\\{{\vec E}_z} = \left( {0.314\frac{{\rm{N}}}{{\rm{C}}}} \right)\hat k\\\end{array}$

Ans:

The electric field vector at P is $\left( {0.314\frac{{\rm{N}}}{{\rm{C}}}} \right)\hat k$.