Question

The minimum surface charge density on the plate, so that a body of mass $$2\text{kg/}{\text{m}}^{\text{2}}$$ may just be lifted, is
A. $$2.84\times {10}^{-5}\text{C/}{\text{m}}^{\text{2}}$$
B. $$2.25\times {10}^{-5}\text{C/}{\text{m}}^{\text{2}}$$
C. $$1.86\times {10}^{-5}\text{C/}{\text{m}}^{\text{2}}$$
D. None of these

Answer 1

Hint: Use the formula for the force per unit area of a charged conductor. Use the concept that this force per unit area of the plate is balanced by the weight per unit area of the body. This gives the relation between the surface charge density on the plate, mass per unit area of the body and permittivity of free space.

Formula used:
The force per unit area $${\displaystyle \frac{F}{ds}}$$ of a charged conductor is given by
$${\displaystyle \frac{F}{ds}}={\displaystyle \frac{{\sigma }^2}{2{\epsilon }_0}}$$ …… (1)
Here, $$\sigma$$ is the surface charge density and $${\epsilon }_0$$ is the permittivity of free space.

Complete answer: or Complete step by step answer:
The mass per unit area of the body is $$2\text{kg/}{\text{m}}^{\text{2}}$$ and the body is just lifted.
$$m=2\text{kg/}{\text{m}}^{\text{2}}$$
Hence, the surface per unit area $${\displaystyle \frac{F}{ds}}$$ of the plate is balanced by the weight per unit area $$mg$$ of the body.
$${\displaystyle \frac{F}{ds}}=mg$$
Here, is the mass per unit area of the body.
The permittivity of free space $${\epsilon }_0$$ is $$8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$$.
$${\epsilon }_0=8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$$
Determine the surface charge density on the plate.
Substitute for $${\displaystyle \frac{F}{ds}}$$ in the above equation.
$${\displaystyle \frac{{\sigma }^2}{2{\epsilon }_0}}=mg$$
Rearrange the above equation for the square of surface charge density $${\sigma }^2$$ on the plate.
$${\sigma }^2=2{\epsilon }_{0}mg$$
Take square root on both sides of the above equation.
$$\sigma =\sqrt{2{\epsilon }_{0}mg}$$
Substitute $$8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$$ for $${\epsilon }_0$$, $$2\text{kg/}{\text{m}}^{\text{2}}$$ for $$m$$ and $$9.8\text{m/}{\text{s}}^2$$ for $$g$$ in the above equation.
$$\sigma =\sqrt{2\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(2\text{kg/}{\text{m}}^{\text{2}}\right)\left(9.8\text{m/}{\text{s}}^2\right)}$$
$$\Rightarrow \sigma =\sqrt{346.92\times {10}^{-12}}$$
$$\Rightarrow \sigma =18.62\times {10}^{-6}$$
$$\Rightarrow \sigma =1.86\times {10}^{-5}\text{C/}{\text{m}}^{\text{2}}$$
Therefore, the surface charge density on the plate is $$1.86\times {10}^{-5}\text{C/}{\text{m}}^{\text{2}}$$.

So, the correct answer is “Option C”.

Note:
Generally, force on an object is balanced by its weight. Here, the force per unit area is balanced by weight per unit area as the area on both sides of the equation gets cancelled.