Physics, 17.10.2020 17:01 brianmcclain7010
5. In Investigation 2, if everything stays the same, except the diameter of the outer ring is doubled, how does the electric field change?
Answers: 2
![\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}](/tpl/images/0816/2895/33a2e.png)
![E(r)_1 = [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}](/tpl/images/0816/2895/bb2a1.png)
![E(r)_2 = [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}](/tpl/images/0816/2895/0d221.png)
![\frac{E(r)_2}{E(r)_1} = \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}](/tpl/images/0816/2895/55aa0.png)
![\frac{E(r)_2}{E(r)_1} = \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}](/tpl/images/0816/2895/c0bda.png)
![\frac{E(r)_2}{E(r)_1} =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}](/tpl/images/0816/2895/62779.png)
![= E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}](/tpl/images/0816/2895/bb697.png)
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5. In Investigation 2, if everything stays the same, except the diameter of the outer ring is double...
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