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A charged particle having a charge of $ - 2 \times {10^{ - 6}}{\text{C}}$ is placed close to a non-conducting plate having a surface charge density as \[4 \times {10^{ - 6}}{\text{C}}{{\text{m}}^{ - 2}}.\] What will be the force of attraction between them:
(A) $0.5\;{\text{N}}$
(B) $0.45\;{\text{N}}$
(C) $0.25\;{\text{N}}$
(D) $0.75\;{\text{N}}$

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Last updated date: 11th Jun 2024
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Answer
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Hint: The condition given in the 'placed close to sheet' question makes the question very simple. The charge is distributed over the sheet's surface. An electric field will be created by this charge distribution and whenever a charge is placed in its magnetic field, depending on the nature of the charge, will affect the charge.
Formula Used: We will use the following formula to find the result
$E = \dfrac{\sigma }{{2{\varepsilon _o}}}$
And
\[F = qE\]
Where
$\sigma $ is the surface charge density
\[q\] is the charge of the particle

Complete Step-by-Step Solution:
Having a uniformly distributed charge for a thin, non-conducting sheet creates an electrical field equal to \[\dfrac{\sigma }{{2{\varepsilon _o}}}.\] If a charge is placed in the electrical field, a force will be experienced. Note that, regardless of the distance, the electrical field is constant. Therefore, the distance or separation of the two need not be known.
The following information is provided to us in the question
$\sigma = 4 \times {10^{ - 6}}{\text{C}}{{\text{m}}^{ - 2}}$
$q = - 2 \times {10^{ - 6}}{\text{C}}$
${\varepsilon _{\text{o}}} = 8.854 \times {10^{ - 12}}{\text{F}}{{\text{m}}^{ - 1}}$
Now we know that
\[F = qE\]
Also $E = \dfrac{\sigma }{{2{\varepsilon _o}}}$
So, we substitute the value of $E$ in the first equation to get
\[F = q\dfrac{\sigma }{{2{\varepsilon _{\text{o}}}}}\]
Now, we will substitute all the known values to find the solution
$F = 2 \times {10^{ - 6}} \times \dfrac{{4 \times {{10}^{ - 6}}}}{{2 \times 8.854 \times {{10}^{ - 12}}}}$
Upon solving, we get
\[\therefore F = 0.45 N\]

Hence, the correct option is (B.)

Note: You also need to know that if the plate is non-conducting when solving this kind of problem, then only we can keep the charge at the place as it is here otherwise if the plate would have conducting then the charge would be present on the surface or far from it depending on it sign as the same signs repel while opposite signs attract each other.