Question

5. In Investigation 2, if everything stays the same, except the diameter of the outer ring is doubled, how does the electric field change?

Answer 1

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

There is a change in the electric field by this factor $\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}$

Explanation:

From the question we are told that

  The electric field is  $E(r)_1 =  [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}$

    Now when the outer diameter is doubled, the radius(b) is also doubled

So

    $E(r)_2 =  [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}$

Now

      $\frac{E(r)_2}{E(r)_1}  =  \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}$

=>   $\frac{E(r)_2}{E(r)_1}  =  \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}$

=>  $\frac{E(r)_2}{E(r)_1}  =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}$

$=> E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}$

Here we see that the electric field changes by the factor $\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}$