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Chapter 21: Problem 43

An electric field is given by \(\vec{E}=E_{0}(y / a) \hat{k},\) where \(E_{0}\) and \(a\) are constants. Find the flux through the square in the \(x\) -y plane bounded by the points \((0,0),(0, a),(a, a),(a, 0)\).

Short Answer

Expert verified
The flux through the square is \(\frac{E_0 a^2}{2}\)

Step by step solution

01

Defining the uniform area vector

The area of the square is given by \(a^2\) and is perpendicular to the electric field direction. So, the area vector will be \(\vec{A} = a^2 \hat{k}\).
02

Applying the formula for Electric Flux

The flux through the square can be found by integrating the electric field over the area. The formula for electric flux is \(Φ = \int \vec{E} \cdot d\vec{A}\), where \(d\vec{A}\) is a small area element. Since the electric field varies linearly with y, we must express the area differential \(d\vec{A}\) in terms of y. The area of the square can also be interpreted as the integral of a line (length a) swept along the y-axis (from 0 to a). Thus, \(d\vec{A} = a dy \hat{k}\).
03

Calculating the Electric Flux

Substitute the expressions for the electric field and the differential area into the flux formula and solve the integral: \(Φ = \int_0^a E_0 \frac{y}{a} \cdot a dy = \frac{E_0}{a} \int_0^a y dy = \frac{E_0 a^2}{2}\), since the integral of \(y\) from 0 to \(a\) is \(\frac{a^2}{2}\).

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Most popular questions from this chapter

The volume charge density inside a solid sphere of radius \(a\) is \(\rho=\rho_{0} r / a,\) where \(\rho_{0}\) is a constant. Find (a) the total charge and (b) the electric field strength within the sphere, as a function of distance \(r\) from the center.

Under what conditions can the electric flux through a surface be written as \(E A\), where \(A\) is the surface area?

A solid sphere \(10 \mathrm{cm}\) in radius carries a \(40-\mu \mathrm{C}\) charge distributed uniformly throughout its volume. It's surrounded by a concentric shell \(20 \mathrm{cm}\) in radius, also uniformly charged with \(40 \mu \mathrm{C}\). Find the electric field (a) \(5.0 \mathrm{cm},\) (b) \(15 \mathrm{cm},\) and (c) \(30 \mathrm{cm}\) from the center.

The field of an infinite charged line decreases as \(1 / r .\) Why isn't this a violation of the inverse-square law?

The electric field in a certain region is given by \(\vec{E}=a x \hat{\imath},\) where \(a=40 \mathrm{N} / \mathrm{C} \cdot \mathrm{m}\) and \(x\) is in meters. Find the volume charge density in the region. (Hint: Apply Gauss's law to a cube 1 m on a side.)
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