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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential
Electric potential is an important concept in electrostatics. It refers to the amount of electric potential energy per unit charge at a specific point in an electric field. Think of it as a measure of how much work would be needed to bring a positive charge from infinity to that point.
For example, in the exercise, point P has an electric potential of 10V, while point Q has a potential of -4V. This means that more work is required to move a positive charge to point P than to point Q, due to the higher potential.
  • Electric potential is measured in volts (V).
  • A positive potential indicates that work is needed to move a positive charge to that point.
  • A negative potential indicates a decrease in potential energy; work is done by the system.
Understanding electric potential helps us calculate the work needed to move charges in an electric field.
Work Done
Work done in electrostatics relates to moving a charge between two points with different potentials. The formula for work done is given by:\[W = q(V_B - V_A)\]where \(W\) is the work done, \(q\) is the charge moved, \(V_A\) is the initial potential, and \(V_B\) is the final potential.
In the context of the problem above, moving 100 electrons from point P to point Q involves calculating the work based on the potential difference.
  • Work is positive when the charge moves to a higher potential.
  • Work is negative when the charge moves to a lower potential, as in this exercise.

By calculating the work done, we determine the energy required or released during the movement of charges.
Electron Charge
An understanding of the electron charge is crucial when dealing with electrostatics problems. An electron has a fundamental charge of \(1.6 \times 10^{-19} C\). This small charge value is significant because it affects how we describe the movement of electrons.
For the problem, this charge was used to compute the total charge moved from point P to Q.
  • The charge of 100 electrons is \(q = 100 \times 1.6 \times 10^{-19} C = 1.6 \times 10^{-17} C\).
  • Electron charge is negative, indicating that electrons naturally move toward positive potentials.
Understanding electron charge is fundamental to quantifying electrostatic phenomena involving electron movement.
Potential Difference
Potential difference, often called voltage, is the difference in electric potential between two points. It's vital in determining the work done on a charge.
In the exercise, the potential difference between points P and Q is computed to decide how much work is done in moving the electrons.

Calculating the potential difference:\[ V_B - V_A = (-4V) - (10V) = -14V \]- **Key Points:**
  • Potential difference indicates the direction and magnitude of energy change during charge movement.
  • A negative potential difference reveals that moving charges from P to Q releases energy.
In electrostatics, knowing the potential difference helps predict how a charge will move and whether energy will be absorbed or released.

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Most popular questions from this chapter

The potential at a point \(\mathrm{x}\) (measured in \(\mu \mathrm{m}\) ) due to some charges situated on the \(\mathrm{x}\) -axis is given by \(\mathrm{V}(\mathrm{x})=\left[(20) /\left(\mathrm{x}^{2}-4\right)\right]\) Volt. The electric field at \(\mathrm{x}=4 \mu \mathrm{m}\) is given by (A) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (B) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (C) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (D) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction

Three identical spheres each having a charge \(\mathrm{q}\) and radius \(R\), are kept in such a way that each touches the other two spheres. The magnitude of the electric force on any sphere due to other two is \(\ldots \ldots \ldots\) (A) \((\mathrm{R} / 2)\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{5} / 4)(\mathrm{q} / \mathrm{R})^{2}\) (B) \(\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{2} / 3)(\mathrm{q} / \mathrm{R})^{2}\) (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{3} / 4)(\mathrm{q} / \mathrm{R})^{2}\) (D) \(-\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{3} / 2)(\mathrm{q} / \mathrm{R})^{2}\)

A thin spherical conducting shell of radius \(\mathrm{R}\) has a charge q. Another charge \(Q\) is placed at the centre of the shell. The electrostatic potential at a point p a distance \((\mathrm{R} / 2)\) from the centre of the shell is ..... (A) \(\left[(q+Q) /\left(4 \pi \epsilon_{0}\right)\right](2 / R)\) (B) \(\left[\left\\{(2 Q) /\left(4 \pi \epsilon_{0} R\right)\right\\}-\left\\{(2 Q) /\left(4 \pi \epsilon_{0} R\right)\right]\right.\) (C) \(\left[\left\\{(2 Q) /\left(4 \pi \in_{0} R\right)\right\\}+\left\\{q /\left(4 \pi \epsilon_{0} R\right)\right]\right.\) (D) \(\left[(2 \mathrm{Q}) /\left(4 \pi \epsilon_{0} \mathrm{R}\right)\right]\)

An electric dipole is placed along the \(\mathrm{x}\) -axis at the origin o. \(\mathrm{A}\) point \(P\) is at a distance of \(20 \mathrm{~cm}\) from this origin such that OP makes an angle \((\pi / 3)\) with the x-axis. If the electric field at P makes an angle \(\theta\) with the x-axis, the value of \(\theta\) would be \(\ldots \ldots \ldots\) (A) \((\pi / 3)+\tan ^{-1}(\sqrt{3} / 2)\) (B) \((\pi / 3)\) (C) \((2 \pi / 3)\) (D) \(\tan ^{-1}(\sqrt{3} / 2)\)

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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