{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 2-solutions

Homework 2-solutions - gonzales(pag757 – Homework 2 –...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: gonzales (pag757) – Homework 2 – Hoffmann – (56745) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod 6 . 1 cm long is uniformly charged and has a total charge of- 15 μ C. Determine the magnitude of the elec- tric field along the axis of the rod at a point 36 . 1231 cm from the center of the rod. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 1 . 04057 × 10 6 N / C. Explanation: Let : ℓ = 6 . 1 cm , Q =- 15 μ C , r = 36 . 1231 cm , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . For a rod of length ℓ and linear charge density (charge per unit length) λ , the field at a dis- tance d from the end of the rod along the axis is E = k e integraldisplay d + ℓ d λ x 2 dx = k e- λ x vextendsingle vextendsingle vextendsingle vextendsingle d + ℓ d = k e λ ℓ d ( ℓ + d ) , where dq = λ dx . The linear charge density (if the total charge is Q ) is λ = Q ℓ so that E = k e Q ℓ ℓ d ( ℓ + d ) = k e Q d ( ℓ + d ) . In this problem, we have the following situa- tion (the distance r from the center is given): d l r r The distance d is d = r- ℓ 2 = 36 . 1231 cm- 6 . 1 cm 2 = 0 . 330731 m , and the magnitude of the electric field is E = k e Q d ( ℓ + d ) = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × |- 1 . 5 × 10- 5 C | (0 . 330731 m)(0 . 061 m + 0 . 330731 m) = 1 . 04057 × 10 6 N / C . Now, the direction must be toward the rod, since the charge distribution is negative (a positive test charge would be attracted). So the sign should be positive, according to the convention stated in the problem. keywords: 002 10.0 points A 9 . 6 g piece of Styrofoam carries a net charge of- . 5 μ C and floats above the center of a very large horizontal sheet of plastic that has a uniform charge density on its surface. The acceleration of gravity is 9 . 8 m / s 2 and the permittivity of free space is 8 . 85419 × 10- 12 C 2 / N / m 2 . What is the charge per unit area on the plastic sheet? Correct answer:- 3 . 33201 μ C / m 2 . Explanation: Let : m = 9 . 6 g , q =- . 5 μ C , g = 9 . 8 m / s 2 , and ǫ = 8 . 85419 × 10- 12 C 2 / N / m 2 . gonzales (pag757) – Homework 2 – Hoffmann – (56745) 2 The field due to a nonconducting infinite sheet of charge is the same as that very close to any plane uniform charge distribution. The field is E = σ 2 ǫ...
View Full Document

{[ snackBarMessage ]}

Page1 / 5

Homework 2-solutions - gonzales(pag757 – Homework 2 –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online