Electric field due to a plane thin charged sheet, charged conducting plate, Practice Problems, FAQs

When a conducting plate is charged, it creates an electric field around it. The electric field as produced by the plate can be useful in many cases such as in parallel plate capacitors, used in energy storage. In case of an infinitely large plate or if the point under consideration is too close to the plate, the field intensity can be considered as uniform. If the plate is positively charged the field lines will emerge out from the plate, parallel to the area vector of the plate. In case the plate is negatively charged, the electric field lines will enter into the plate, anti-parallel to the area vector of the plate. Let’s visualize and calculate these electric fields here.

Table of content:

Calculation of electric field due to a plane thin charged sheet
Calculation of electric field due to a thick charged conducting sheet
Practice Problems
FAQs

Calculation of electric field due to a plane thin charged sheet

When we know the electric field around a regular shaped object, we can calculate the contained charge in it. It can be achieved by the application of Gauss Law.

According to the Gauss law,

We know, $ε_{0} = 8.854 \times 10^{- 12} F m^{- 1}$

We can also do the opposite. Let’s see how!

Consider a flat thin sheet which is comparatively infinite in size having surface charge density . As the sheet is infinite it will have an electric field with same magnitude but in opposite directions at any two points equidistant from the sheet on the opposite sides. To calculate the electric field, we will draw a Gaussian surface in a cylindrical shape with ends on the two sides of cross-sectional area S. Electrical field lines will be tangential to the curved surface of the cylindrical Gaussian surface and at the two ends, these electrical field lines will be parallel to the area vector.

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$E (2 S) = \frac{σ S}{ε_{0}}$

$\Rightarrow E = \frac{σ}{{2 ε}_{0}}$

Note: This result is correct for a real charged sheet if the points under consideration are not near the edges and the distance from the sheet is small compared to the dimension of the sheet so that we can consider this as an infinite plane thin sheet of charges.

Calculation of electric field due to a thick charged conducting sheet

If charge is provided to a conducting plate, the charge will distribute itself on the entire outer surface. If the plate is of uniform thickness and of infinite size, the surface charge density will be uniform and will be the same on both sides.

According to the Gauss law,

$\Rightarrow \frac{σ 2 S}{ε_{0}} = E (2 S) i . e . E = \frac{σ}{ε_{0}}$

Note: E would have been inward in case the charge contained in it was negative.

Thus, the field due to a charged conducting plate is twice the field due to an infinite thin plane sheet of charges for a given charge density.

Practice Problems:

Q. A large thin charged sheet has surface charge density 3 Cm-2. What will be the electric field intensity at a small distance from the sheet?

A. The electric field at a small distance due to a plane thin sheet of charge is

$E = \frac{σ}{{2 ε}_{0}} = \frac{3}{2 \times 8.854 \times 10^{- 12}} \approx 1.7 \times 10^{11} N C^{- 1}$

Q. A large thick conducting charged plate has surface charge density 17.7 Cm-2. What will be the electric field intensity at a small distance from the charged plate?

A. Electric field due to a thick charged conducting surface is,

$E = \frac{σ}{ε_{0}} = \frac{17.7}{8.854 \times 10^{- 12}} \approx 2 \times 10^{12} N C^{- 1}$

Q. In calculating the electric field of a charged large sheet or a charged plate by Gauss’s law, can we take the point where the field is to be measured far from the sheet or plate?

A. No.

In application of Gauss's law and deducing the electric field we consider the electric flux to be uniform throughout. This can only be achieved if the point is near the surface and the surface is large enough, so that the flux distribution remains uniform.

Q. A conducting plate of dimension 1 m×1 m is charged by 17 C. Another thin sheet of dimension 1 m×1 m is charged by 17 C. What will be the electric field intensity at a point 1 mm from the mid of the plate and sheet?

A. The cases can be considered as an infinitely large plate and infinitely large thin sheet respectively.

For the conducting plate, $σ = \frac{17}{2 (1 \times 1)} = 8.5 C m^{- 2}$ and the electric field intensity $E = \frac{σ}{ε_{0}} = \frac{8.5}{8.854 \times 10^{- 12}} = 9.6 \times 10^{11} N C^{- 1}$

For the thin sheet, $σ = \frac{17}{1 \times 1} = 17 C m^{- 2}$ and the electric field intensity $E = \frac{σ}{2 ε_{0}} = \frac{17}{2 \times 8.854 \times 10^{- 12}} = 9.6 \times 10^{11} N C^{- 1}$

FAQs:

Q. Can we only take a cylindrical gaussian surface for calculation of electric field due to a plane thin charged sheet?
A. We can take a cube or cuboid also, since direction of electric field will be either parallel or perpendicular to the surface on this case also and performing the integration would still be easy.

Q. What is the field inside a thick conducting sheet?
A. The field inside a thick conducting sheet will be zero since the field inside a conductor is always zero in electrostatic condition.

Q. There is a large conducting charged plate of surface charge density . What is the net electric flux for an area S of the sheet?
A. The electric flux will be, $E 2 S = \frac{σ 2 S}{ε_{0}}$

Here the factor 2 comes because of the two faces of the charged plate.

Q. There is a large conducting charged plate of surface charge density . What is the electric intensity at a point close to the surface?
A. The electric field intensity at a point close to the surface will be, $E = \frac{σ}{ε_{0}}$