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Chapter 23: Problem 28

(II) Point a is 26 \(\mathrm{cm}\) north of a \(-3.8 \mu \mathrm{C}\) point charge, and point \(\mathrm{b}\) is 36 \(\mathrm{cm}\) west of the charge (Fig. \(27 ) .\) Determine \((a) V_{\mathrm{b}}-V_{\mathrm{a}},\) and \((b) \hat{\mathbf{E}}_{\mathrm{b}}-\vec{\mathbf{E}}_{\mathrm{a}}\) (magnitude and direction).

Short Answer

Expert verified
The potential difference is calculated using \(V=\frac{kQ}{r}\) and the electric field difference using \(E=\frac{k|Q|}{r^2}\). Directions affect signs.

Step by step solution

01

Understand the Problem

We are given a point charge of \(-3.8 \mu C\). We need to determine the potential difference \((V_b - V_a)\) between two points, \(a\) and \(b\), which are located 26 cm north and 36 cm west of the charge, respectively. We also need to calculate the differences in the electric field magnitude and direction at these two points.
02

Convert units

Convert the distances from cm to meters for calculation purposes. Point \(a\)'s position is 26 cm or 0.26 m north of the charge. Point \(b\)'s position is 36 cm or 0.36 m west of the charge.
03

Calculate Potential Difference Vb - Va

The potential \(V\) at a distance \(r\) from a point charge \(Q\) is given by \(V = \frac{kQ}{r}\) where \(k \approx 8.99 \times 10^9 \) N m虏/C虏. Compute necessary potentials: - \(V_a = \frac{k \times (-3.8 \times 10^{-6})}{0.26}\)- \(V_b = \frac{k \times (-3.8 \times 10^{-6})}{0.36}\)The potential difference \(V_b - V_a = V_b - V_a\).
04

Calculate Electric Fields at Points a and b

The electric field \(E\) at a distance \(r\) from a point charge \(Q\) is given by \(E = \frac{k|Q|}{r^2}\). Compute electric fields:- \(E_a = \frac{k \times 3.8 \times 10^{-6}}{(0.26)^2}\)- \(E_b = \frac{k \times 3.8 \times 10^{-6}}{(0.36)^2}\).
05

Determine Direction of Electric Fields

The direction of the electric field caused by a negative point charge is radially towards the charge. Therefore, - \(E_a\) is directed south (opposite to north) - \(E_b\) is directed east (opposite to west).
06

Calculate Magnitude of Electric Field Difference

The difference in electric field magnitude is\[ |\vec{E}_b| - |\vec{E}_a| \ = |E_b| - |E_a| \ = \frac{k \times 3.8 \times 10^{-6}}{(0.36)^2} - \frac{k \times 3.8 \times 10^{-6}}{(0.26)^2} \].
07

Final Result and Interpretation

After calculating the above steps, substitute the values for \(k\), \(Q\), and the distances to find both \(V_b - V_a\) and \(|\hat{E}_b| - |\vec{E}_a|\). The signs and magnitudes will provide information about the relative field strengths and potential differences between the two points.

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Key Concepts

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

Electric Field
The electric field is a region around a charged particle where other charges experience a force. It's a vector field, meaning it has both magnitude and direction. The electric field \( E \) around a point charge \( Q \) is given by the formula \( E = \frac{k |Q|}{r^2} \). Here, \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \) N m虏/C虏, and \( r \) is the distance from the charge.
  • The magnitude of the electric field decreases with the square of the distance.
  • A positive charge has an electric field pointing away, while a negative one points toward it.
In the example provided, - \( E_a \) is calculated at a distance of 0.26 m north of the charge.- \( E_b \) at 0.36 m west.
These positions result in different magnitudes of the electric field at these points, showing how sensitive the electric field is to distance changes. Electric fields from negative charges always point toward the source charge, dictating the field's direction.
Point Charge
A point charge refers to a hypothetical charge located at a single point in space. It is used to simplify electrical problems by considering charges as having no dimensions. In many scenarios, like the one described, point charges are used to calculate fields and potentials.
  • It assumes the charge's size is negligible compared to the distances involved.
  • Point charges carry either a positive or negative charge and have a corresponding electric field.
In the exercise, the point charge is \(-3.8 \mu C\) (microcoulombs). The negative sign indicates that the charge is an electron-donor type, meaning it attracts positive charges and repels negative ones. Calculating forces, fields, or potentials with point charges often uses the aforementioned formulae, allowing us to predict behavior in electrical circuits and static electricity scenarios.
Potential Difference
Potential difference, often referred to as voltage, is the difference in electric potential between two points in space. It represents the work needed to move a unit charge from one point to another within an electric field. The formula to find the potential, \( V \), due to a point charge is \( V = \frac{kQ}{r} \), making potential inversely proportional to the distance from the charge.
  • Potential difference tells us the energy conversion rate per unit charge moving between two points.
  • It's measured in volts, where one volt is one joule per coulomb.
In the context provided, you were tasked with finding \( V_b - V_a \), the potential difference between points \( a \) and \( b \). - Point \( a \) is closer to the charge, leading to a higher potential compared to \( b \), which is further away.- Therefore, the calculation involves determining the potential at each point using their respective distances and the charge value, then subtracting to find the difference.Understanding potential difference provides insight into how electric current will move, since it naturally flows from higher to lower potential areas.

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Most popular questions from this chapter

(II) The liquid-drop model of the nucleus suggests that highenergy oscillations of certain nuclei can split ("fission") a large nucleus into two unequal fragments plus a few neutrons. Using this model, consider the case of a uranium nucleus fissioning into two spherical fragments, one with a charge \(q_{1}=+38 e\) and radius \(r_{1}=5.5 \times 10^{-15} \mathrm{~m},\) the other with \(q_{2}=+54 e\) and \(r_{2}=6.2 \times 10^{-15} \mathrm{~m} .\) Calculate the electric potential energy \((\mathrm{MeV})\) of these fragments, assuming that the charge is uniformly distributed throughout the volume of each spherical nucleus and that their surfaces are initially in contact at rest. The electrons surrounding the nuclei can be neglected. This electric potential energy will then be entirely converted to kinetic energy as the fragments repel each other. How does your predicted kinetic energy of the fragments agree with the observed value associated with uranium fission (approximately \(200 \mathrm{MeV}\) total \() ?\left[1 \mathrm{MeV}=10^{6} \mathrm{eV}\right.\)

(III) You are trying to determine an unknown amount of charge using only a voltmeter and a ruler, knowing that it is either a single sheet of charge or a point charge that is creating it. You determine the direction of greatest change of potential, and then measure potentials along a line in that direction. The potential versus position (note that the zero of position is arbitrary, and the potential is measured relative to ground) is measured as follows: $$ \begin{array}{lllllllllll} \hline x(\mathrm{~cm}) & 0.0 & 1.0 & 2.0 & 3.0 & 4.0 & 5.0 & 6.0 & 7.0 & 8.0 & 9.0 \\ \mathrm{~V} \text { (volts) } & 3.9 & 3.0 & 2.5 & 2.0 & 1.7 & 1.5 & 1.4 & 1.4 & 1.2 & 1.1 \\ \hline \end{array} $$ (a) Graph V versus position. Do you think the field is caused by a sheet or a point charge? \((b)\) Graph the data in such a way that you can determine the magnitude of the charge and determine that value. \((c)\) Is it possible to determine where the charge is from this data? If so, give the position of the charge.

A nonconducting sphere of radius \(r_{2}\) contains a concentric spherical cavity of radius \(r_{1}\). The material between \(r_{1}\) and \(r_{2}\) carries a uniform charge density \(\rho_{\mathrm{E}}\left(\mathrm{C} / \mathrm{m}^{3}\right) .\) Determine the electric potential \(V,\) relative to \(V=0\) at \(r=\infty,\) as a function of the distance \(r\) from the center for \((a) r>r_{2},\) (b) \(r_{1}

In a photocell, ultraviolet (UV) light provides enough energy to some electrons in barium metal to eject them from the surface at high speed. See Fig. \(36 .\) To measure the maximum energy of the electrons, another plate above the barium surface is kept at a negative enough potential that the emitted electrons are slowed down and stopped, and return to the barium surface. If the plate voltage is \(-3.02 \mathrm{V}\) (compared to the barium) when the fastest electrons are stopped, what was the speed of these electrons when they were emitted?

(I) How much work does the electric field do in moving a proton from a point with a potential of \(+185 \mathrm{~V}\) to a point where it is \(-55 \mathrm{~V} ?\)
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