# Concentric Cylindrical Conducting Shells

## Homework Statement

The picture of the problem can be found here: http://www.2shared.com/photo/U_JIkDks/Capture.html

The questions that I'm having trouble with are:

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

C = Q / ΔV
E = λ / 2∏rε
λ = Q/L

## The Attempt at a Solution

(a) Alright for the first one I'm saying that the new capacitance will just equal the initial capacitance (before the charge densities were changed) because capacitance only depends on the geometry of the capacitor. If lambda is increased then that means the charges (Q) was increased, this increases the electric field which by this formula ΔV = Ed also increases the potential difference. Since the potential difference doubled, the total charge doubled (λ doubles) so the capacitance remains the same since it is only a ratio between the charge and potential difference.

(b) From above since potential difference doubled then λ must double as well so my new λ is 0.82μC/m

Does this seem correct? Thanks

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rude man
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## Homework Statement

The picture of the problem can be found here: http://www.2shared.com/photo/U_JIkDks/Capture.html
The questions that I'm having trouble with are:

(a) The magnitudes of the charge densities on the inner and outer shells are now changed (keeping λinner = -λouter) so that the resulting 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?
Correct.

(b) What is λouter,new ?

C = Q / ΔV
E = λ / 2∏rε
λ = Q/L

## The Attempt at a Solution

(a) Alright for the first one I'm saying that the new capacitance will just equal the initial capacitance (before the charge densities were changed) because capacitance only depends on the geometry of the capacitor. If lambda is increased then that means the charges (Q) was increased, this increases the electric field which by this formula ΔV = Ed also increases the potential difference. Since the potential difference doubled, the total charge doubled (λ doubles) so the capacitance remains the same since it is only a ratio between the charge and potential difference.

(b) From above since potential difference doubled then λ must double as well so my new λ is 0.82μC/m

Does this seem correct? Thanks
Yes. Referring to a unit length of the cable, Q = CV, C hasn't changed so since V has doubled, so has Q.