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Question: A charged particle having a charge \( - 2 \times {10^{ - 6}}C\) is placed close to the non-conductin...

A charged particle having a charge −2×10−6C - 2 \times {10^{ - 6}}C is placed close to the non-conducting plate having a surface charge density 4×10−6Cm−24 \times {10^{ - 6}}C{m^{ - 2}}. The force of attraction between the particle and the plate is nearly:
A. 0.9N
B. 0.71N
C. 0.62N
D. 0.45N

Explanation

Solution

In this problem you have to use the formula of force between a charge particle due to the electric field and for that you need to find the electric field due to the plate using the formula and the plate is non conducting having surface charge density (σ\sigma ). Doing this will solve your problem and will give you the right answer.

Complete answer:
We know that force F on the charge particle q due electric field E is F = qE.
We also know that the electric field due to the infinite plate is E = σ2ε0\dfrac{\sigma }{{2{\varepsilon _0}}}.
It has been given that σ=4×10−6Cm−2\sigma = 4 \times {10^{ - 6}}C{m^{ - 2}} and q=−2×10−6q = - 2 \times {10^{ - 6}}
So we can calculate electric field as,
E = σ2ε0=4×10−6Cm−22×9×10−12=2.23×1061=2.23×106Nm−1\dfrac{\sigma }{{2{\varepsilon _0}}} = \dfrac{{4 \times {{10}^{ - 6}}C{m^{ - 2}}}}{{2 \times 9 \times {{10}^{ - 12}}}} = \dfrac{{2.23 \times {{10}^6}}}{1} = 2.23 \times {10^6}N{m^{ - 1}}.
Then force F = −2×10−6×0.223×106=−0.446N - 2 \times {10^{ - 6}} \times 0.223 \times {10^6} = - 0.446N
The force is -0.446 N which can be written as -0.45 N.
Negative sign indicates that the force is attractive.
So, the magnitude of the force is 0.45N.

So, the correct answer is “Option D”.

Note:
When you get to solve such problems you need to know the formulas like E = σ2ε0\dfrac{\sigma }{{2{\varepsilon _0}}}, F = qE and their respective units and then you also need to know that if the plate is non conducting then only we can keep the charge at the place as it is here otherwise if the plate would have been conducting then the charge would be present on the surface or far away from it depending upon it sign as same signs repel while opposite signs attracts each other. The formula E = σ2ε0\dfrac{\sigma }{{2{\varepsilon _0}}} is only for one side of the plate for both sides of the infinite plate the formula of electric field is σε0\dfrac{\sigma }{{{\varepsilon _0}}} that is also called electric field due to sheet. Knowing this will help you and solve your problem.