Questions 6-8 A nonconducting rod of length 2L is placed along the x-axis such that its midpoint is at the origin. The rod is positively charged and has a non-uniform linear charge density given by X(x) = A₁x/L for L ≤ x ≤ L with A₁ > 0 being a positive constant. k = (47€)¹ is the electric constant. 6. What is the total charge of the rod? (a) A₁L/2 (b) 2λ₁L (c) λ₁L/4 (d) A₁L (e) 4X₁L 7. What is the electric potential, due to the charged rod, at a point (0, y) located on the positive y-axis? (a) 2kλ1₁(L − y)/L__ (b) 2kλ₁ (√L² + y² −y) /L_(c) 2kλ₁/√ (d) 2kλ₁(√√Zy²/L - y)/L (e) 0 8. Use your answer to the previous question to find the electric field at the point (x, y) = (0, L). (a) o (b) k(A₁/√√2)/Lĵ (c) kλ₁(2-√2)/Lĵ (d) k(λ₁/L) ĵ (e) kλ₁ (2√2-1)/Lĵ

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Questions 6-8
A nonconducting rod of length 2L is placed along the x-axis such that its midpoint is at the origin. The rod is positively
charged and has a non-uniform linear charge density given by X(x) = A₁x/L for L ≤ x ≤ L with A₁ > 0 being a
positive constant. k = (47€o)-¹ is the electric constant.
6. What is the total charge of the rod?
(a) A₁L/2 (b) 2λ₁L (c) λ₁L/4 (d) λ₁L (e) 4X₁L
7. What is the electric potential, due to the charged rod, at a point (0, y) located on the positive y-axis?
(a) 2kλ₁(L − y)/L (b) 2kλ₁ (√√I² + y² −y) /L (c) 2kλ₁/√ (d) 2kλ₁ (√2y²/L - y)/L (e) 0
8. Use your answer to the previous question to find the electric field at the point (x, y) = (0, L).
(a) ♂ (b) k(A₁/√√2)/Lĵ (c) kλ₁(2-√2)/Lĵ (d) k(λ₁/L) ĵ (e) kλ₁ (2√2-1)/Lĵ
Transcribed Image Text:Questions 6-8 A nonconducting rod of length 2L is placed along the x-axis such that its midpoint is at the origin. The rod is positively charged and has a non-uniform linear charge density given by X(x) = A₁x/L for L ≤ x ≤ L with A₁ > 0 being a positive constant. k = (47€o)-¹ is the electric constant. 6. What is the total charge of the rod? (a) A₁L/2 (b) 2λ₁L (c) λ₁L/4 (d) λ₁L (e) 4X₁L 7. What is the electric potential, due to the charged rod, at a point (0, y) located on the positive y-axis? (a) 2kλ₁(L − y)/L (b) 2kλ₁ (√√I² + y² −y) /L (c) 2kλ₁/√ (d) 2kλ₁ (√2y²/L - y)/L (e) 0 8. Use your answer to the previous question to find the electric field at the point (x, y) = (0, L). (a) ♂ (b) k(A₁/√√2)/Lĵ (c) kλ₁(2-√2)/Lĵ (d) k(λ₁/L) ĵ (e) kλ₁ (2√2-1)/Lĵ
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