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Chapter 11: Problem 1520

When a Piece of Polythene is rubbed with wool, a charge of \(-2 \times 10^{-7}\) is developed on polythene. The mass transferred to polythene is $\ldots \mathrm{kg}$. (A) \(11.38 \times 10^{-19}\) (B) \(5.69 \times 10^{-19}\) (C) \(2.25 \times 10^{-19}\) (D) \(9.63 \times 10^{-19}\)

Short Answer

Expert verified
The mass transferred to polythene is (A) \(1.138 \times 10^{-18}\ \mathrm{kg}\).

Step by step solution

01

Calculate the number of electrons transferred

We know that the charge of a single electron is approximately \(e = -1.6 \times 10^{-19} \mathrm{C}\). We're given that the charge developed on polythene is \(Q = -2 \times 10^{-7} \mathrm{C}\). To find the number of electrons transferred, we can use the formula: Number of electrons transferred = \(\frac{\text{Total Charge}}{\text{Charge of one electron}}\)
02

Calculate the mass transferred

First, find the number of electrons transferred by dividing the total charge (Q) by the charge of one electron (e): Number of electrons transferred = \(\frac{-2 \times 10^{-7}}{-1.6 \times 10^{-19}}\) Number of electrons transferred = \(1.25 \times 10^{12}\) Now, we need to find the mass of electrons transferred. We know that the mass of a single electron is approximately \(m_e = 9.1 \times 10^{-31} \mathrm{kg}\). To find the mass transferred, we can multiply the number of electrons transferred by the mass of a single electron: Mass transferred = Number of electrons transferred \(\times\) Mass of one electron
03

Calculate the mass transferred

Calculate the mass transferred by multiplying the number of electrons transferred by the mass of one electron: Mass transferred = \(1.25 \times 10^{12} \times 9.1 \times 10^{-31}\) Mass transferred = \(1.1375 \times 10^{-18}\) Finally, we can compare our result with the given options: (A) \(11.38 \times 10^{-19} \approx 1.138 \times 10^{-18}\) (correct) (B) \(5.69 \times 10^{-19}\) (C) \(2.25 \times 10^{-19}\) (D) \(9.63 \times 10^{-19}\) The correct answer is (A) \(1.138 \times 10^{-18}\ \mathrm{kg}\).

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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

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A Semicircular rod is charged uniformly with a total charge \(\mathrm{Q}\) coulomb. The electric field intensity at the centre of curvature is $\ldots \ldots \ldots$ (A) \(\left[(2 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (B) \(\left[(3 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (C) \(\left[(\mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (D) \(\left[(4 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\)

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A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}\(. The separation between its plates is \)\mathrm{d}$. The space between the plates is now filled with two dielectrics. One of the dielectric constant \(\mathrm{K}_{1}=3\) and thickness \(\mathrm{d} / 3\) while the other one has dielectric constant \(\mathrm{K}_{2}=6\) and thickness \(2 \mathrm{~d} / 3\). Capacitance of the capacitor is now (A) \(1.8 \mathrm{pF}\) (B) \(20.25 \mathrm{pF}\) (C) \(40.5 \mathrm{pF}\) (D) \(45 \mathrm{pF}\)
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