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Chapter 22: Problem 23

An electric field is given by \(\vec{E}=E_{0} \hat{\jmath},\) where \(E_{0}\) is a constant. Find the potential as a function of position, taking \(V=0\) at \(y=0\)

Short Answer

Expert verified
The electric potential as a function of position is \(V(y) = - E_{0} y\). It indicates that the potential decreases linearly with increasing y, in the opposite direction of the electric field.

Step by step solution

01

Set up the integral for electric potential

Electric potential, V, at position y is given by the negative integral of the electric field from a reference point to the point y, as the electric potential is the work done against the electric field. Here, the expression for electric potential is \(V(y) = - \int_{0}^{y} \vec{E} \cdot d\vec{y}\).
02

Substitute the value of the electric field

Next, substitute the value of \(\vec{E} = E_{0} \hat{\jmath}\) into the integral: \(V(y) = - \int_{0}^{y} E_{0} \hat{\jmath} \cdot d\vec{y}\). As \(\hat{\jmath} \cdot d\vec{y} = dy\), the integral simplifies to \(V(y) = - \int_{0}^{y} E_{0} dy\).
03

Evaluate the integral

Evaluate the integral over the interval from 0 to y:\(V(y) = - [E_{0} y]_{0}^{y}\), since the integral of \(E_{0} = E_{0}y + C\), where C is the constant of integration. Substituting the limits of integration we get: \(V(y) = - E_{0} y\). The integration constant C disappears, as the potential is set to 0 at \(y=0\).

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Most popular questions from this chapter

The potential at the surface of a 10 -cm-radius sphere is \(4.8 \mathrm{kV}\) What's the sphere's total charge, assuming charge is distributed in a spherically symmetric way?

The electric potential in a region is \(V=-V_{0}(r / R),\) where \(V_{0}\) and \(R\) are constants and \(r\) is the radial distance from the origin. Find expressions for the magnitude and direction of the electric field in this region.

An electron passes point \(A\) moving at \(6.5 \mathrm{Mm} / \mathrm{s}\). At point \(B\) it's come to a stop. Find the potential difference \(\Delta V_{A B}\)

Must the electric field be zero at any point where the potential is zero? Explain.

Two points \(A\) and \(B\) lie \(15 \mathrm{cm}\) apart in a uniform electric field, with the path \(A B\) parallel to the field. If the potential difference \(\Delta V_{A B}\) is \(840 \mathrm{V},\) what's the field strength?
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