24.10. A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, s

24.10. A cylindrical capacitor consists of a solid inner...

Key Concepts

Cylindrical Capacitor Geometry

This concept involves understanding the physical configuration of a capacitor formed by two coaxial cylinders. It is important to grasp how the inner and outer radii, along with the length, define the spatial boundaries within which the electric field exists. The separation between the conductors (often filled with air or another dielectric) is key to determining the capacitor’s properties.

Capacitance Formula for Coaxial Cylindrical Capacitors

The capacitance of a cylindrical capacitor is derived from the potential difference created by a charge distribution on the inner cylinder and its effect on the outer cylinder. The formula, often expressed as C = (2???L) / ln(b/a) where ?? is the permittivity of free space, L is the length of the capacitor, and a and b are the inner and outer radii respectively, highlights the logarithmic dependency on the ratio of the radii. This relationship is central to solving problems involving cylindrical capacitors.

Relationship Between Charge, Capacitance, and Voltage

This key concept involves the fundamental equation Q = CV, which relates the charge stored in a capacitor (Q) to its capacitance (C) and the potential difference across its plates (V). Understanding this relationship is crucial when determining either the amount of charge stored or the potential difference for a given capacitance.

Charge Per Unit Length

In systems with extended geometries, like a cylindrical capacitor, it is often useful to compute charge per unit length (?). This concept involves normalizing the total charge over the length of the capacitor to facilitate analysis and comparison independent of the overall size of the system. It reinforces the idea of distributed charge in linear systems.

Transcript

00:01 In the given problem there is a cylindrical capacitor having two coadgeal cylinders like this.

00:18 The radius of the inner cylinder means a has been given as 0 .250 centimeter or we can say this is 0 .250 into 10 based power minus 2 meter the radius of this outer cylinder is missing which we have to find the length of this whole of the arrangement in the capacitor is given to be 12 .0 centimeter or this is 12 .0 into 10 dashpar minus 2 meter the capacitance the magnitude of capacitance of this capacitor cylindrical capacitor has been given as 26 .7 pico ferret or we can say this is 36 .7 into 10 -dhspar minus 12 ferret.

01:27 In the first part of the problem we have to find the magnitude of this outer radius for which we will use the expression for the capacitance of such a cylindrical capacitor which is nothing but 2 pi epsilon not into l divided by the natural logarithm of the ratio of these two radii.

01:50 So using that, first of all, we will find an expression for this natural logarithm of the ratio of b is to a, which comes out to be 2 pi epsilon knot l divided by the capacitance and putting the known values, all the known values, here, it becomes the natural logarithm of b by a, it is a is equal to 1 by 2 into 9 into 10 dash 18 into 9 for the value of this 2 pi epsilon not then into length which is length of the capacitor which is given as 0 .1 2 after converting this 12 into 10 per minus 2 meter finally in the division in the denominator this is the capacitance which is 36 .7 into 10 dash power minus 12 so then we solve all these things here we get the this value here comes out to be 0 .182 so now we will convert this natural logarithm into log at this so to do this we multiply this natural logarithm by 2 .303 so now it can be written as log at this 10 of b by a is equal to 0 .182 or we can convert it into log at this 10 of b by a is equal to 0 .182 divided by this 2 .303 which finally comes out to be 0 .07888 so the value of this b by a is nothing but the antilog of 0 .07888 and this antilog comes out to be 1 .199 or approximately we can say this is 1 .0 .0 .2 .2 .2 so so the radius of outer cylinder here becomes b is equal to a into 1 .2 meter or we can say this is a which was 0 .2 250 into 10 dash bar minus 2 meter into 1 .2 in meters.

04:40 So this answer finally comes out to be if we find it in centimeter the radius of the outer cylinder is 0 .5.

04:49 3 0 .30.

04:53 So this is the answer for the first part of the problem...

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