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Chapter 11: Problem 1570

Two Points \(P\) and \(Q\) are maintained at the Potentials of \(10 \mathrm{v}\) and \(-4 \mathrm{v}\), respectively. The work done in moving 100 electrons from \(\mathrm{P}\) to \(\mathrm{Q}\) is \(\ldots \ldots \ldots\) (A) \(2.24 \times 10^{-16} \mathrm{~J}\) (B) \(-9.60 \times 10^{-17} \mathrm{~J}\) (C) \(-2.24 \times 10^{-16} \mathrm{~J}\) (D) \(9.60 \times 10^{-17} \mathrm{~J}\)

Short Answer

Expert verified
The work done in moving 100 electrons from point P to point Q is \(-2.24 \times 10^{-16} J\).

Step by step solution

01

Identify the given values

We are given the potentials of point P as \(10V\) and point Q as \(-4V\). We are moving 100 electrons from P to Q.
02

Use the work done formula

The work done in moving a charge between two points in a potential field is given by the formula: \[W = q(V_B - V_A)\] Here, W: work done, q: charge being moved, \(V_A\): potential of point A (initial point), \(V_B\): potential of point B (final point). In this case, point P is the initial point and point Q is the final point. So, \(V_A = 10V\), \(V_B = -4V\), q: charge of 100 electrons.
03

Calculate the charge q

We know that one electron has a charge of \(1.6 \times 10^{-19} C\). So, for 100 electrons, the total charge q is: \[q = 100 \times 1.6 \times 10^{-19} C = 1.6 \times 10^{-17} C\]
04

Compute the work done W

Now, substitute the values we have obtained into the work done formula: \[W = (1.6 \times 10^{-17} C)(-4V - 10V)\] \[W = (1.6 \times 10^{-17} C)(-14V)\] \[W = -2.24 \times 10^{-16} J\] Therefore, the work done in moving 100 electrons from point P to point Q is \(-2.24 \times 10^{-16} J\), which is option (C).

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Most popular questions from this chapter

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed \(10 \mathrm{~cm}\) apart in air. The work done to bring them $4 \mathrm{~cm}$ closer is (A) Zero (B) \(4.8 \mathrm{~J}\) (C) \(3.5 \mathrm{~J}\) (D) \(-5.8 \mathrm{~J}\)

The circular plates \(\mathrm{A}\) and \(\mathrm{B}\) of a parallel plate air capacitor have a diameter of \(0.1 \mathrm{~m}\) and are $2 \times 10^{-3} \mathrm{~m}\( apart. The plates \)\mathrm{C}\( and \)\mathrm{D}$ of a similar capacitor have a diameter of \(0.1 \mathrm{~m}\) and are $3 \times 10^{-3} \mathrm{~m}\( apart. Plate \)\mathrm{A}\( is earthed. Plates \)\mathrm{B}$ and \(\mathrm{D}\) are connected together. Plate \(\mathrm{C}\) is connected to the positive pole of a \(120 \mathrm{~V}\) battery whose negative is earthed, The energy stored in the system is (A) \(0.1224 \mu \mathrm{J}\) (B) \(0.2224 \mu \mathrm{J}\) (C) \(0.4224 \mu \mathrm{J}\) (D) \(0.3224 \mu \mathrm{J}\)

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) \(-4 \times 10^{3} \mathrm{E}_{0}\)

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) \(4 \times 10^{-2} \mathrm{~J}\)

A charged particle of mass \(1 \mathrm{~kg}\) and charge \(2 \mu \mathrm{c}\) is thrown from a horizontal ground at an angle \(\theta=45^{\circ}\) with speed $20 \mathrm{~m} / \mathrm{s}\(. In space a horizontal electric field \)\mathrm{E}=2 \times 10^{7} \mathrm{~V} / \mathrm{m}$ exist. The range on horizontal ground of the projectile thrown is $\ldots \ldots \ldots$ (A) \(100 \mathrm{~m}\) (B) \(50 \mathrm{~m}\) (C) \(200 \mathrm{~m}\) (D) \(0 \mathrm{~m}\)
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