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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

An inclined plane making an angle of \(30^{\circ}\) with the horizontal is placed in an uniform electric field \(E=100 \mathrm{Vm}^{-1}\). A particle of mass \(1 \mathrm{~kg}\) and charge \(0.01 \mathrm{c}\) is allowed to slide down from rest from a height of \(1 \mathrm{~m} .\) If the coefficient of friction is \(0.2\) the time taken by the particle to reach the bottom is $\ldots \ldots . .$ sec (A) \(2.337\) (B) \(4.337\) (C) 5 (D) \(1.337\)

Point charges \(q_{1}=2 \mu c\) and \(q_{2}=-1 \mu c\) care kept at points \(\mathrm{x}=0\) and \(\mathrm{x}=6\) respectively. Electrical potential will be zero at points ..... (A) \(\mathrm{x}=-2, \mathrm{x}=2\) (B) \(\mathrm{x}=1, \mathrm{x}=5\) (C) \(\mathrm{x}=4, \mathrm{x}=12\) (D) \(\mathrm{x}=2, \mathrm{x}=9\)

N identical drops of mercury are charged simultaneously to 10 volt. when combined to form one large drop, the potential is found to be 40 volt, the value of \(\mathrm{N}\) is \(\ldots \ldots\) (A) 4 (B) 6 (C) 8 (D) 10

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) $\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]$ (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) $\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]$ (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

A thin spherical shell of radius \(R\) has charge \(Q\) spread uniformly over its surface. Which of the following graphs, figure most closely represents the electric field \(\mathrm{E}\) (r) produced by the shell in the range $0 \leq \mathrm{r}<\infty\(, where \)\mathrm{r}$ is the distance from the centre of the shel1.
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