Introduction to Electrodynamics
Introduction to Electrodynamics
4th Edition
ISBN: 9781108420419
Author: David J. Griffiths
Publisher: Cambridge University Press
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Chapter 2.5, Problem 2.48P
To determine

The potential difference between north pole and center of an inverted hemispherical bowl.

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For problem 11 of the text, calculate the potential in pixpoV at a point r = 0.398 R from the center of the sphere. (5 sig figs)
There is a surface charge ρS=2.9 [nC/m2] in the plane 6*x+10*y+9*z+-4=0. If Va=27 at point A(-1,9,1) find the Scalar Potential Vb [V] at point B(-4,6,3
When an uncharged conducting sphere of radius a is placed at the origin of an xyz coordinate system that lies in an initially uniform electric field É = E,k, the resulting elec- tric potential is V(x, y, 2) = V, for points inside the sphere and E,a°z V(x, y, 2) = V – E,z + (x² + y² + z²)3/2 for points outside the sphere, where V, is the (constant) elec- tric potential on the conductor. Use this equation to deter- mine the x, y, and z components of the resulting electric field (a) inside the sphere and (b) outside the sphere.

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Introduction to Electrodynamics

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