Textbook Question
What is the capacitance of the two metal spheres shown in FIGURE EX26.22?
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26. Capacitors & Dielectrics
Capacitors & Capacitance
5:17 minutesProblem 50c
Textbook Question
A cylindrical capacitor (Example 24–2) has Ra = 3.5 mm and Rb = 0.50 mm. The two conductors have a potential difference of 625 V, with the inner conductor at the higher potential. Calculate the electric field at the surface of the inner conductor.
Verified step by step guidance1
Step 1: Understand the problem. The electric field at the surface of the inner conductor of a cylindrical capacitor can be calculated using the formula for the electric field in a cylindrical geometry. The formula is derived from Gauss's law and is given by: , where V is the potential difference, r is the radius at which the field is being calculated, Rₐ is the outer radius, and R₆ is the inner radius.
Step 2: Identify the given values. From the problem, the potential difference V = 625 V, the inner radius R₆ = 0.50 mm = 0.0005 m, and the outer radius Rₐ = 3.5 mm = 0.0035 m. The electric field needs to be calculated at the surface of the inner conductor, so r = R₆.
Step 3: Substitute the values into the formula. Replace V, r, Rₐ, and R₆ in the formula: . This will give the electric field at the surface of the inner conductor.
Step 4: Simplify the denominator. Calculate the natural logarithm term , which represents the ratio of the outer radius to the inner radius.
Step 5: Perform the division and logarithmic calculation. Once the denominator is simplified, divide the potential difference by the product of the radius and the logarithmic term to find the electric field. Ensure units are consistent throughout the calculation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Cylindrical Capacitor
A cylindrical capacitor consists of two concentric cylindrical conductors separated by an insulating material. The capacitance depends on the radii of the conductors and the length of the cylinder. The electric field within the capacitor is influenced by the potential difference between the conductors and the geometry of the system.
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Electric Field
The electric field (E) is a vector field that represents the force per unit charge experienced by a positive test charge placed in the field. For a cylindrical capacitor, the electric field can be calculated using Gauss's law, which relates the electric field to the charge enclosed by a Gaussian surface. The electric field is directed radially outward from the inner conductor.
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Gauss's Law
Gauss's Law states that the electric flux through a closed surface is proportional to the enclosed electric charge. It is mathematically expressed as ∮E·dA = Q_enc/ε₀, where E is the electric field, dA is the differential area vector, Q_enc is the enclosed charge, and ε₀ is the permittivity of free space. This law is particularly useful for calculating electric fields in symmetric charge distributions, such as cylindrical capacitors.
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