Question

Two very large, non conducting plastic sheets, each \(10.0\,cm\) thick, carry uniform charge densities \({\sigma _1}\), \({\sigma _2}\), \({\sigma _3}\) and \({\sigma _4}\) on their surfaces (
Fig. E22.30). These surface charge densities have the values \({\sigma _1} = - 6.00\,\mu C/{m^2}\) , \({\sigma _2} = + 5.00\,\mu C/{m^2}\), \({\sigma _3} = + 2.00\,\mu C/{m^2}\), \({\sigma _4} = + 4.00\,\mu C/{m^2}\). Use Gauss law to find the magnitude and direction of the electric field at the following points, far from the edges of these sheets: (a) point A, \(5.00\,cm\)from the left face of the left hand sheet; (b) Point B, \(1.25\,cm\) from the inner surface of the right hand sheet; (c) point C, in the middle of the right-hand sheet.

Short answer

1. The magnitude of the electric field at point \(A\) is \(2.82 \times 1{0^5}\,N/C\) towards left.

Step-by-step solution

Identification of given data

• The surface charge density is given by,
  \({\sigma _1} = - 6.00\,\mu C/{m^2}\), \({\sigma _2} = + 5.00\,\mu C/{m^2}\), \({\sigma _3} = + 2.00\,\mu C/{m^2}\), \({\sigma _4} = + 4.00\,\mu C/{m^2}\)
• It is given that the point A, \(5.00\,cm\)from the left face of the left hand sheet

Understanding the concept:

The electric field intensity is defined as the force per unit charge. Mathematically it can be written as,

\(\overrightarrow E = \frac{{\overrightarrow F }}{Q}\) ….. (1)

Here\(\overrightarrow E \)is the electric field intensity vector,\(\overrightarrow F \)is the force vector, and\(Q\)is the magnitude of the charge.

The electric field intensity due to infinite charge sheet is given by,

\(E = \frac{\sigma }{{2{\varepsilon _o}}}\)…… (2)

Here \(E\) is the magnitude of the electric field, \(\sigma \) is the surface charge density, \({\varepsilon _o}\) is the permittivity of free space (vacuum).

Determine the magnitude and direction of the electric field at the point A:

Let us assume that the electric field sue to surface charge density\({\sigma _1}\), \({\sigma _2}\), \({\sigma _3}\)and\({\sigma _4}\)is\({E_1}\),\({E_2}\),\({E_3}\)and\({E_4}\).

Then the net electric field at pint\(A\)is ,

\(E = {E_1} - {E_2} - {E_3} - {E_4}\)

Using the equation (2), we can write

\(E = \frac{{{\sigma _1}}}{{2{\varepsilon _o}}} - \frac{{{\sigma _2}}}{{2{\varepsilon _o}}} - \frac{{{\sigma _3}}}{{2{\varepsilon _o}}} - \frac{{{\sigma _4}}}{{2{\varepsilon _o}}}\)

Substitute\( - 6.00\,\mu C/{m^2}\)for\({\sigma _1}\),\( + 5.00\,\mu C/{m^2}\)for\({\sigma _2}\),\( + 2.00\,\mu C/{m^2}\)for\({\sigma _3}\),\( + 4.00\,\mu C/{m^2}\)for\({\sigma _4}\)and\(8.85 \times 1{0^{ - 12}}\,F/m\)for\({\varepsilon _o}\)into the above equation,

\(\begin{aligned}E = \frac{{ - 6 - 5 - 2 - 4}}{{8.85 \times 1{0^{ - 12}}}} \times 1{0^{ - 6}}\\ = - 2.82 \times 1{0^5}\,N/C\end{aligned}\)

Therefore the magnitude of the electric field at point\(A\)is\(2.82 \times 1{0^5}\,N/C\)towards left.

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