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
`(ln[(b)/(a) ])/(ln[(2b)/(a) ])`

Step-by-step explanation:

From the question we are told that

The electric field is
`E(r)_1 =  [(V_o)/(ln(b) -ln(a)) ] * (1)/(r)`

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

So

`E(r)_2 =  [(V_o)/(ln(2b) -ln(a)) ] * (1)/(r)`

Now

`(E(r)_2)/(E(r)_1)  =  ((V_o)/(ln(b) -ln(a)) ] * (1)/(r))/((V_o)/(ln(2b) -ln(a)) ] * (1)/(r))`

=>
`(E(r)_2)/(E(r)_1)  =  (V_o)/(ln(b) -ln(a)) ] * (1)/(r) * ( ln(2b) -ln(a))/(V_o) ] * (r)/(1)`

=>
`(E(r)_2)/(E(r)_1)  =(ln[(b)/(a) ])/(ln[(2b)/(a) ])`

`=> E(r)_2 =(ln[(b)/(a) ])/(ln[(2b)/(a) ]) }{E(r)_1}`

Here we see that the electric field changes by the factor
`(ln[(b)/(a) ])/(ln[(2b)/(a) ])`