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## GR0877 Problem 90

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) For $a one has $\displaystyle \varphi(r) = -\int_a^r E dr = -\int_a^r \frac{Q}{4 \pi \epsilon_0} \frac{dr}{r^2} = \frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{a} \right)$; for $r>b$ $\displaystyle \varphi(r) = -\int_a^b E dr - \int_b^r E dr = -\int_a^b \frac{Q}{4 \pi \epsilon_0} \frac{dr}{r^2} - 0 = \frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{b} - \frac{1}{a} \right)$.

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Categories: 0877 Tags: , , , ,
1. 12.11.2011 at 03:37

For problems like this with two parts of the solution, solve for the part of the solution with the most variance in answers. for this one you immediately get the answer when you solve for Region II, whereas if you solve for Region I you have four to choose from and no choice but to waste more time.

2. 24.09.2014 at 19:36

All except D are discontinuous at r=b, so you can solve this one immediately using elimination. b also shouldn’t show up in region 1 equations.

3. 23.10.2015 at 00:07

Noting that the region ii value should be a nonzero constant leaves D and E. Obviously b shouldn’t show up in the region i equation which leaves D.

4. 28.02.2017 at 12:28

Why doesn’t potential go to zero if we go to infinity?

• 15.10.2017 at 03:43

They set the zero at the surface of the in er shell at radius a. Normally we do set V_inf = 0 but they changed it up here.