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liq [111]
4 years ago
12

The magnitudes of the charge densities on the inner and outer shells are now changed (keeping λinner = -λouter) so that the resu

lting potential difference doubles (Vca,new = 2Vca,initial). How does Cnew, the capacitance of a one meter length of the system of conductors when the charge density is changed, compare to C, the initial capacitance of a one meter length of the system of conductors? (b) What is λouter,new ?
Physics
1 answer:
elena-14-01-66 [18.8K]4 years ago
5 0

Answer: a) Cnew=Cinitial ; b) λouter new=  2*λ outer initial

Explanation: In order to explain this question we have to take into account the expression of teh cylinder capacitor given by:

C/L= (2*π*εo)/ln (b/a)=  where b and a are the outer and inner radius, respectively. L is the length of the capacitor.

As you can se this formule depents  of geometrical characateristics  of the capacitor.

The capacitance is the same after change the densities of charge.

On the other hand,

The new charge in each cylinder ( inner and outer) is determined

The new potential is 2 times the initial one so

V new= 2* Vinitial

Also we know that

Vnew= Q/C= λnew*L/C;   C= constant

using this formule and considering  that V new is doubled then the charge per one meter length,  is also doubled .

This is as follow:

Vnew=  λnew*L/C=  

λnew =  (2*Vinitial)* C/L= 2  (λ initial)

Then  λouter new = 2* λouter initial

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Answer: Option B

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Where,

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                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

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When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

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