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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

A charge \(\mathrm{Q}\) is divided into two parts and then they are placed at a fixed distance. The force between the two charges is always maximum when the charges are \(\ldots \ldots\) (A) \((Q / 3),(Q / 3)\) (B) \((\mathrm{Q} / 2),(\mathrm{Q} / 2)\) (C) \((Q / 4),(3 Q / 4)\) (D) \((Q / 5),(4 Q / 5)\)

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be $\ldots . .$ (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

A variable condenser is permanently connected to a \(100 \mathrm{~V}\) battery. If capacitor is changed from \(2 \mu \mathrm{F}\) to \(10 \mu \mathrm{F}\). then energy changes is equal to (A) \(2 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(6.5 \times 10^{-2} \mathrm{~J}\) (D) \(4 \times 10^{-2} \mathrm{~J}\)

The inward and outward electric flux for a closed surface in units of \(\mathrm{Nm}^{2} / \mathrm{C}\) are respectively \(8 \times 10^{3}\) and $4 \times 10^{3}\(. Then the total charge inside the surface is \)\ldots \ldots \ldots \ldots . . \mathrm{c}$. (A) \(\left[\left(-4 \times 10^{3}\right) / \epsilon_{0}\right]\) (B) \(-4 \times 10^{3}\) (C) \(4 \times 10^{3}\) (D) \(-4 \times 10^{3} \mathrm{E}_{0}\)

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q\). The potential difference between the centers of the two rings is $\ldots .$ (A) 0 (B) $\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]$ (C) $\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.$ (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)
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