{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw04 - Physics 21 Spring 2011 This problem can be solved...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 21 Spring, 2011 Solution to HW-4 23-5 In Figure 23-29, a proton is a distance d/ 2 directly above the center of a square of side d . What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge d .) Electric flux through a surface is defined by the integral Φ E = integraltext E · d A . The brute force method of solving this problem would be to write out an expression for E as a function of position on the surface of the square, and integrate that expression. But the hint tells us to use symmetry to solve the problem. If we imagine the proton is located in the center of a cube with edge d , then it is in a position with a great deal of symmetry: it is a distance d/ 2 from the center of each face the cube. We conclude that the flux through each face of the cube must be equal to 1 / 6 of the total flux through the cube. But how do we calculate the total flux through the cube? The cube is totally enclosing the proton. An interesting property of flux is that the total flux through any enclosing surface is the same. We saw in Lecture 3 that a sphere is the easiest surface to calculate the total flux due to a point charge. The result of that calculation is Φ E = q ǫ 0 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}