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An electron is placed halfway between the two cylinders (i.e. r =), I. H. the charge is released from rest, which surface is it accelerated towards? ) What is the electric potential at r, relative to the surface that the electron eventually strikes? Hint: Consider the result from part c., with the appropriate substitution.

An electron is placed halfway between the two cylinders (i.e. r =), I. H. the charge is released from rest, which surface is it accelerated towards? ) What is the electric potential at r, relative to the surface that the electron eventually strikes? Hint: Consider the result from part c., with the appropriate substitution.

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15 mm, and R₂ = 2 cm. assume |2|= 1, p = 5C, R₁ = 5 mm, R₁ = d). An electron is placed halfway between the two cylinders (i.e. r = +). ii. the charge is released from rest, which surface is it accelerated towards? ) What is the electric potential at r, relative to the surface that the electron eventually strikes? Hint: Consider the result from part c., with the appropriate substitution.
Transcribed Image Text:15 mm, and R₂ = 2 cm. assume |2|= 1, p = 5C, R₁ = 5 mm, R₁ = d). An electron is placed halfway between the two cylinders (i.e. r = +). ii. the charge is released from rest, which surface is it accelerated towards? ) What is the electric potential at r, relative to the surface that the electron eventually strikes? Hint: Consider the result from part c., with the appropriate substitution.
An infinitely long, solid insulating cylinder with radius Ra is placed concentric within a conducting cylindrical shell of inner radius R, and outer radius Re. The inner cylinder has a uniform volume charge density +[pl, and the outer cylinder has a net linear charge density of -3121. Assume Ip Rål> 132 for all parts. a. b. C. Sketch the scenario, indicating how much charge is in each region / surface. What are the charge densities on the inner and outer surfaces of the conducting cylinder? Part a. may be made easier by providing your answer in terms A' = pRa (the 'linear charge density' of the inner cylinder), instead of p. alculating the electric field using Gauss' Law. Let r denote the radial distance from the center of the insulating cylinder. To solve Gauss' law in all space, there are four (4) regions to consider. The result for the electric field E(r) in regions l- III are given below Ē(r) = E₁(r) = -rf 2€0 PR²1 Ē₁ (r) = f 201 Em (r) = 0 I claim Avab is I claim Ey (r) is Using Gauss' law, derive the expression for E(r) in region IV, where r > Rc. r< Ra Make a statement on the term in parenthesis in En(1). Ra <r <Ro Rp <r < Re 3λ Ew (1) = DR (1 - 32²) f 2E0 r What is the electric potential difference between the inner surface of the conducting cylinder and the outer surface of the insulating cylinder (in region II)? Hint: See the result from part b for Ēn (r). AVab= V(R₂) - V(Ra)= PRZ 2€0 log Explicitly point out in each step which terms in the derivation contribute negative signs.
Transcribed Image Text:An infinitely long, solid insulating cylinder with radius Ra is placed concentric within a conducting cylindrical shell of inner radius R, and outer radius Re. The inner cylinder has a uniform volume charge density +[pl, and the outer cylinder has a net linear charge density of -3121. Assume Ip Rål> 132 for all parts. a. b. C. Sketch the scenario, indicating how much charge is in each region / surface. What are the charge densities on the inner and outer surfaces of the conducting cylinder? Part a. may be made easier by providing your answer in terms A' = pRa (the 'linear charge density' of the inner cylinder), instead of p. alculating the electric field using Gauss' Law. Let r denote the radial distance from the center of the insulating cylinder. To solve Gauss' law in all space, there are four (4) regions to consider. The result for the electric field E(r) in regions l- III are given below Ē(r) = E₁(r) = -rf 2€0 PR²1 Ē₁ (r) = f 201 Em (r) = 0 I claim Avab is I claim Ey (r) is Using Gauss' law, derive the expression for E(r) in region IV, where r > Rc. r< Ra Make a statement on the term in parenthesis in En(1). Ra <r <Ro Rp <r < Re 3λ Ew (1) = DR (1 - 32²) f 2E0 r What is the electric potential difference between the inner surface of the conducting cylinder and the outer surface of the insulating cylinder (in region II)? Hint: See the result from part b for Ēn (r). AVab= V(R₂) - V(Ra)= PRZ 2€0 log Explicitly point out in each step which terms in the derivation contribute negative signs.
Expert Solution
Step 1

Given:

The linear charge density is, λ=1 nCcm.

The surface charge density is, ρ=5 nCcm3.

The radius, Ra=5 mm.

The radius, Rb=15 mm.

The radius, Rc=2 cm.

 

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