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Chapter 25: Q. 69 (page 713)

FIGURE P25.69 shows a thin rod of length L and charge Q. Find an expression for the electric potential a distance x away from the center of the rod on the axis of the rod.

Short Answer

Expert verified

The electric potential is given as $V = \frac{k Q}{L} I n \frac{2 x + L}{2 x - L}$

Step by step solution

01

Given Information

We need to find an expression for the electric potential a distance $x$ away from the center of the rod on the axis of the rod.

02

Simplify

Consider the fact that the potential is a scalar quantity, which means

$V = k \frac{q}{r}$

We can express this as an integral. In our case, instead of the charge, we'll have the linear charge density $Q / L$ , and the integration variable, let it be $l$ . The potential at point $x$ will therefore be given as

$V = k {}_{\frac{- L}{2}}^{\frac{L}{2}}{\frac{Q / L}{l + x}} d l$

Solving this integral,

$V = \frac{k Q}{L} {}_{- L / 2}^{L / 2}{\frac{d l}{l + x}} = \frac{k Q}{L} {[\ln (l + x)]}_{- L / 2}^{L / 2} = \frac{k Q}{L} [\ln (x + L / 2) - \ln (x_{L} / 2]$

By the properties of the logarithm and by simplifying with the factor of two, we get

$V = \frac{k Q}{L} I n \frac{2 x + L}{2 x - L}$

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Most popular questions from this chapter

$F I G U R E Q 25.11$ shows three points near two point charges. The charges have equal magnitudes. For each part, rank in order, from most positive to most negative, the potentials $V_{a}$ to $V_{c}$ .

Two positive point charges are $5.0 cm$ apart. If the electric potential energy is $72 μJ$ , what is the magnitude of the force between the two charges?

$F I G U R E Q 25.9$ shows two points inside a capacitor. Let $V = 0 V$ at the negative plate.
$(a)$ What is the ratio $\frac{V_{2}}{V_{1}}$ of the electric potentials? Explain.
$(b)$ What is the ratio $\frac{E_{2}}{E_{1}}$ of the electric field strengths?

What is the speed of an electron that has been accelerated from rest through a potential difference of $1000 V$ ?

An electric dipole has dipole moment p. If r W s, where

s is the separation between the charges, show that the electric

potential of the dipole can be written

$V = \frac{1}{4 π \in \circ} \frac{p \cos θ}{r^{2}}$

where r is the distance from the center of the dipole and u is the

angle from the dipole axis.

See all solutions

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