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Chapter 23: Problem 43

An electric field is established in a nonuniform rod. A voltmeter is used to measure the potential difference between the left end of the rod and a point a distance \(x\) from the left end. The process is repeated, and it is found that the data are described by the relationship \(\Delta V=270 x^{2},\) where \(\Delta V\) has the units \(\mathrm{V} / \mathrm{m}^{2}\). What is the \(x\) -component of the electric field at a point \(13 \mathrm{~cm}\) from the left end?

Short Answer

Expert verified
Answer: The x-component of the electric field at a point 13 cm from the left end is -7.02 × 10² V/m.

Step by step solution

01

Write down the given relationship

We are given the relationship between the potential difference \(\Delta V\) and the distance \(x\) as \(\Delta V = 270x^2\), where \(\Delta V\) has units of V/m².
02

Recall the relationship between electric field and potential

The relationship between the electric field \(E\) and the potential difference \(\Delta V\) is given by \(\Delta V = -\int E dx\), where the \(x\)-component of the electric field is \(E_x\).
03

Differentiate the potential difference with respect to x

To find the electric field at a specific point, we need to differentiate the potential difference with respect to distance \(x\). Let \(\frac{d(\Delta V)}{dx} = \frac{d(270x^2)}{dx}\). Differentiating \(270x^2\) with respect to \(x\), we get \(\frac{d(\Delta V)}{dx} = 540x\).
04

Find the electric field at the given point

Now, we can find the \(x\)-component of the electric field at the given point by using the result from step 3 as the \(x\)-component of the electric field is given by the negative of the derivative of potential difference with respect to x: \(E_x = -\frac{d(\Delta V)}{dx}\). At a point 13 cm from the left end, \(x = 13 \times 10^{-2}\) meters. Plug this value into our result from step 3: \(E_x = -540x = -540(13 \times 10^{-2}) = -7.02 \times 10^2 \mathrm{V/m}\) The \(x\)-component of the electric field at a point \(13 \mathrm{~cm}\) from the left end is \(-7.02 \times 10^2 \mathrm{V/m}\).

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Most popular questions from this chapter

Two protons at rest and separated by \(1.00 \mathrm{~mm}\) are released simultaneously. What is the speed of either at the instant when the two are \(10.0 \mathrm{~mm}\) apart?

Two fixed point charges are on the \(x\) -axis. A charge of \(-3.00 \mathrm{mC}\) is located at \(x=+2.00 \mathrm{~m}\) and a charge of \(+5.00 \mathrm{mC}\) is located at \(x=-4.00 \mathrm{~m}\) a) Find the electric potential, \(V(x),\) for an arbitrary point on the \(x\) -axis. b) At what position(s) on the \(x\) -axis is \(V(x)=0 ?\) c) Find \(E(x)\) for an arbitrary point on the \(x\) -axis.

What would be the consequence of setting the potential at \(+100 \mathrm{~V}\) at infinity, rather than taking it to be zero there? a) Nothing; the field and the potential would have the same values at every finite point. b) The electric potential would become infinite at every finite point, and the electric field could not be defined. c) The electric potential everywhere would be \(100 \mathrm{~V}\) higher, and the electric field would be the same. d) It would depend on the situation. For example, the potential due to a positive point charge would drop off more slowly with distance, so the magnitude of the electric field would be less.

A solid conducting sphere of radius \(R\) is centered about the origin of an \(x y z\) -coordinate system. A total charge \(Q\) is distributed uniformly on the surface of the sphere. Assuming, as usual, that the electric potential is zero at an infinite distance, what is the electric potential at the center of the conducting sphere? a) zero c) \(Q / 2 \pi \epsilon_{0} R\) b) \(Q / \epsilon_{0} R\) d) \(Q / 4 \pi \epsilon_{0} R\)

The electric field, \(\vec{E}(\vec{r}),\) and the electric potential \(V(\vec{r}),\) are calculated from the charge distribution, \(\rho(\vec{r}),\) by integrating Coulomb's Law and then the electric field. In the other direction, the field and the charge distribution are determined from the potential by suitably differentiating. Suppose the electric potential in a large region of space is given by \(V(r)=V_{0} \exp \left(-r^{2} / a^{2}\right),\) where \(V_{0}\) and \(a\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) is the distance from the origin. a) Find the electric field \(\vec{E}(\vec{r})\) in this region. b) Determine the charge density \(\rho(\vec{r})\) in this region, which gives rise to the potential and field. c) Find the total charge in this region. d) Roughly sketch the charge distribution that could give rise to such an electric field.
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