91Ó°ÊÓ

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).

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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