
A charged particle having charge $-2\times 10^{-6} C$ is placed close to a non conducting plate having a surface charge density as $4 \times 10^{-6} Cm^{-2}$. What will be the force of attraction between them.
Answer
574.8k+ views
Hint: The condition given in the question ‘ placed close to sheet’ makes the question very easy. The charge is distributed over the surface of the sheet. This charge distribution will create an electric field and whenever a charge is placed in its magnetic field, it will affect the charge, depending upon the nature of charge.
Formula used:
$E = \dfrac{\sigma}{2\epsilon_{\circ}}, F = qE$
Complete answer:
For a thin, non conducting sheet, having uniformly distributed charge creates an electric field equal to $\dfrac{\sigma}{2\epsilon_{\circ}}$. When a charge is placed in the electric field, it will experience a force. Note that the electric field is constant, irrespective of the distance. Hence we need not to know the distance or separation of the two.
Hence using F=qE, and $E= \dfrac{\sigma}{2\epsilon_{\circ}}$, we get;
$F = q \times \dfrac{\sigma}{2\epsilon_{\circ}}$
Now, given $\sigma = 4 \times 10^{-6} Cm^{-2} , q= -2\times 10^{-6} C \ and \ \epsilon_{\circ} = 8.854 \times 10^{-12}Fm^{-1}$
Putting the values in the equation;
$F = 2\times 10^{-6} \times \dfrac{4\times 10^{-6}}{2\times 8.854 \times 10^{-12}} = 0.45 N$
Hence, the net force acting on the charge is 0.45N.
Now since the charge is negative in nature and the sheet contains positive charge distributed, hence the force will be attractive in nature and hence will act towards the sheet.
Note:
One can object here that the formula $E = \dfrac{\sigma}{2\epsilon_{\circ}}$ is only for infinite sheets and not for a plate. Yes that’s true, but in the question, it is clearly mentioned that the charge is close to the sheet and even distance is not given. Hence we will consider that the distance is very less and hence all the electric field is towards the normal of the plate. This is something like this when we place our finger very very close to the eye that it becomes blurry. At that time even a little finger appeared huge.
Formula used:
$E = \dfrac{\sigma}{2\epsilon_{\circ}}, F = qE$
Complete answer:
For a thin, non conducting sheet, having uniformly distributed charge creates an electric field equal to $\dfrac{\sigma}{2\epsilon_{\circ}}$. When a charge is placed in the electric field, it will experience a force. Note that the electric field is constant, irrespective of the distance. Hence we need not to know the distance or separation of the two.
Hence using F=qE, and $E= \dfrac{\sigma}{2\epsilon_{\circ}}$, we get;
$F = q \times \dfrac{\sigma}{2\epsilon_{\circ}}$
Now, given $\sigma = 4 \times 10^{-6} Cm^{-2} , q= -2\times 10^{-6} C \ and \ \epsilon_{\circ} = 8.854 \times 10^{-12}Fm^{-1}$
Putting the values in the equation;
$F = 2\times 10^{-6} \times \dfrac{4\times 10^{-6}}{2\times 8.854 \times 10^{-12}} = 0.45 N$
Hence, the net force acting on the charge is 0.45N.
Now since the charge is negative in nature and the sheet contains positive charge distributed, hence the force will be attractive in nature and hence will act towards the sheet.
Note:
One can object here that the formula $E = \dfrac{\sigma}{2\epsilon_{\circ}}$ is only for infinite sheets and not for a plate. Yes that’s true, but in the question, it is clearly mentioned that the charge is close to the sheet and even distance is not given. Hence we will consider that the distance is very less and hence all the electric field is towards the normal of the plate. This is something like this when we place our finger very very close to the eye that it becomes blurry. At that time even a little finger appeared huge.
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