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Chapter 11: Problem 1561

An oil drop of 12 excess electrons is held stationary under a constant electric field of \(2.55 \times 10^{4} \mathrm{Vm}^{-1}\). If the density of the oil is \(1.26 \mathrm{gm} / \mathrm{cm}^{3}\) then the radius of the drop is \(\ldots \ldots \ldots \mathrm{m}\). (A) \(9.75 \times 10^{-7}\) (B) \(9.29 \times 10^{-7}\) (C) \(9.38 \times 10^{-8}\) (D) \(9.34 \times 10^{-8}\)

Short Answer

Expert verified
The radius of the oil drop is approximately \(9.34 \times 10^{-8} \mathrm{m}\).

Step by step solution

01

Write down the given values

We are given the following values: - number of excess electrons: 12 - electric field strength: \(2.55 \times 10^{4} \mathrm{Vm}^{-1}\) - density of the oil: \(1.26 \mathrm{gm} / \mathrm{cm}^{3}\)
02

Calculate electrostatic force

The electrostatic force (FE) acting on the oil drop can be calculated using the formula: \(F_E = n_exe\cdot E \) where \(n_e\) is the number of excess electrons, \(E\) is the electric field strength and \(e\) is the elementary charge (\(1.6 \times 10^{-19} \mathrm{C}\)). Plugging the values into the formula, we get: \(F_E = 12 \times (1.6 \times 10^{-19} \mathrm{C}) \times (2.55 \times 10^{4} \mathrm{Vm}^{-1})\) \(F_E = 4.896 \times 10^{-14} \mathrm{N}\)
03

Calculate gravitational force

As the oil drop is held stationary, the gravitational force (FG) is equal to the electrostatic force (FE). Using the formula for gravitational force acting on a sphere, we have: \(F_G = \frac{4}{3}\pi r^3 \rho g \) where r is the radius of the oil drop, \(\rho\) is the density of the oil, and g is the gravitational acceleration (\(9.81 \mathrm{m/s^2}\)). Since \(F_G = F_E\), we have: \(\frac{4}{3}\pi r^3 \rho g = 4.896 \times 10^{-14} \mathrm{N}\)
04

Solve for the oil drop radius

Now, we have to solve for the radius (r): \(\frac{4}{3}\pi r^3 (1.26 \times 10^3 \mathrm{kg/m^3}) (9.81 \mathrm{m/s^2}) = 4.896 \times 10^{-14} \mathrm{N}\) \(r^3 = \frac{4.896 \times 10^{-14} \mathrm{N}}{\frac{4}{3}\pi (1.26 \times 10^3 \mathrm{kg/m^3})(9.81 \mathrm{m/s^2})}\) \(r = \sqrt[3]{\frac{4.896 \times 10^{-14} \mathrm{N}}{\frac{4}{3}\pi (1.26 \times 10^3 \mathrm{kg/m^3})(9.81 \mathrm{m/s^2})}}\) \(r \approx 9.34 \times 10^{-8} \mathrm{m}\) Considering the provided options, the correct answer is: (D) \(9.34 \times 10^{-8} \mathrm{m}\)

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Most popular questions from this chapter

The electric Potential \(\mathrm{V}\) at any Point $0(\mathrm{x}, \mathrm{y}, \mathrm{z}\( all in meters \))\( in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}\( volt. The electric field at the point \)(1 \mathrm{~m}, 0.2 \mathrm{~m})\( in volt meter is \)\ldots \ldots .$ (A) 8 , along negative \(\mathrm{x}\) - axis (B) 8 , along positives \(\mathrm{x}\) - axis (C) 16 , along negative \(\mathrm{x}\) -axis (D) 16 , along positives \(\mathrm{x}\) -axis

Electric charges of \(+10 \mu \mathrm{c}, 5 \mu \mathrm{c},-3 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed at the corners of a square of side $\sqrt{2 m}\( the potential at the centre of the square is \)\ldots \ldots$ (A) \(1.8 \mathrm{~V}\) (B) \(1.8 \times 10^{5} \mathrm{~V}\) (C) \(1.8 \times 10^{6} \mathrm{~V}\) (D) \(1.8 \times 10^{4} \mathrm{~V}\)

A parallel plate air capacitor has a capacitance \(\mathrm{C}\). When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be (A) \(200 \%\) (B) \(33.3 \%\) (C) \(400 \%\) (D) \(66.6 \%\)

4 Points charges each \(+q\) is placed on the circumference of a circle of diameter \(2 \mathrm{~d}\) in such a way that they form a square. The potential at the centre is \(\ldots \ldots .\) (A) 0 (B) \((4 \mathrm{kd} / \mathrm{q})\) (C) \((\mathrm{kd} / 4 \mathrm{q})\) (D) \((4 \mathrm{kq} / \mathrm{d})\)

Two identical capacitors have the same capacitance \(\mathrm{C}\). one of them is charged to a potential \(\mathrm{V}_{1}\) and the other to \(\mathrm{V}_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)\) (B) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}\right)\) (C) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}\) (D) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)
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