Question 333: Electrostats & Capacitance

Question

A small bob of mass 100 mg and charge $+10 \mu \mathrm{C}$ is connected to an insulating string of length 1 m . It is brought near to an infinitely long non-conducting sheet of charge density ' $\sigma$ ' as shown in figure. If string subtends an angle of $45^{\circ}$ with the sheet at equilibrium the charge density of sheet will be.
(Given, $\epsilon_0=8.85 	imes 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}$ and acceleration due to gravity, $\mathrm{g}=10 \frac{\mathrm{~m}}{\mathrm{~s}^2}$ )

Accepted Answer

Step 1: Balancing Forces on the Bob
At equilibrium, the electric force ($qE$) pulling the bob toward the sheet is balanced by the weight of the bob ($mg$) pulling it down.
Step 2: Formula for Electric Field Near Sheet
The electric field ($E$) created by a large sheet of charge is given by $E = \dfrac{\sigma}{2\varepsilon_0}$.
Step 3: Replace $E$ in the Force Equation
The equation from earlier becomes: $$q \left[\frac{\sigma}{2 \varepsilon_0}ight] = mg$$
Step 4: Solve for $\sigma$
We solve for $\sigma$ by rearranging: $$\sigma = \frac{2\varepsilon_0 mg}{q}$$
Step 5: Substitute the Values
Put in the given values: $\varepsilon_0 = 8.85 	imes 10^{-12}$ F/m, $m = 100$ mg $= 100 	imes 10^{-6}$ kg, $g = 10$ m/s$^2$, $q = 10 	imes 10^{-6}$ C.
So, $$\sigma = \frac{2 	imes 8.85 	imes 10^{-12} 	imes 100 	imes 10^{-6} 	imes 10}{10 	imes 10^{-6}}$$
Step 6: Calculate
We do the multiplication and division: $$\sigma = 17.7 	imes 10^{-10}~\mathrm{C}/\mathrm{m}^2$$
Step 7: Convert to nC/m$^2$
$1~\mathrm{nC} = 10^{-9}~\mathrm{C}$, so $$\sigma = 1.77~\mathrm{nC}/\mathrm{m}^2$$

Answer

$1.77 \mathrm{~nC} / \mathrm{m}^2$

Answer

$0.885 \mathrm{~nC} / \mathrm{m}^2$

Answer

$885 \mathrm{~nC} / \mathrm{m}^2$

Answer

$17.7 \mathrm{~nC} / \mathrm{m}^2$

Electrostats & Capacitance

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