HW 5 - benavides(jjb2356 – homework 05 – Turner...

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: benavides (jjb2356) – homework 05 – Turner – (59130) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A charged mass on the end of a light string is attached to a point on a uniformly charged vertical sheet of infinite extent. The acceleration of gravity is 9 . 8 m / s 2 and the permittivity of free space is 8 . 854 × 10 − 12 C 2 / N · m 2 . 7 1 . 9 c m θ ˆ ı ˆ . 1 μ C 1 g A r e a l c h a r g e d e n s i t y . 1 5 μ C / m 2 Find the angle θ the thread makes with the vertically charge sheet. Correct answer: 4 . 94015 ◦ . Explanation: Let : g = 9 . 8 m / s 2 , ǫ = 8 . 854 × 10 − 12 C 2 / N · m 2 , m = 1 g = 0 . 001 kg , σ = 0 . 15 μ C / m 2 = 1 . 5 × 10 − 7 C / m 2 , q = 0 . 1 μ C = 1 × 10 − 7 C , and L = 71 . 9 cm = 0 . 719 m . The length L of the string is superfluous. Let the tension in the string be denoted by T . The electric field due to the infinite sheet is constant in the x-direction and is vector E = σ 2 ǫ ˆ ı . In the ˆ ı and ˆ directions, force equilibrium tells us T sin θ = q · σ 2 ǫ T cos θ = m g tan θ = T sin θ T cos θ = q σ 2 m g ǫ θ = arctan parenleftbigg q σ 2 m g ǫ parenrightbigg = arctan bracketleftbigg (1 × 10 − 7 C) (1 . 5 × 10 − 7 C / m 2 ) 2 (0 . 001 kg) (9 . 8 m / s 2 ) ǫ bracketrightbigg = 4 . 94015 ◦ . 002 (part 2 of 2) 10.0 points What value would σ in order for he angle 79 ◦ ? Correct answer: 8 . 9278 μ C / m 2 . Explanation: From the previous part, we know σ = 2 m g tan θ q ǫ = 2 (0 . 001 kg) ( 9 . 8 m / s 2 ) tan(79 ◦ ) 1 × 10 − 7 C · (8 . 854 × 10 − 12 C 2 / N · m 2 ) 10 6 μ C 1 C = 8 . 9278 μ C / m 2 . 003 10.0 points A net positive charge Q is placed on a large, thin conducting plate of are A . In electrostatic equilibrium the charge den- sity σ on each surface and the electric field E outside the plate are 1. σ = Q A , E = σ ǫ 2. σ = 0 , E = σ ǫ 3. σ = Q 2 A , E = σ 2 ǫ 4. σ = Q 2 A , E = 0 5. σ = Q 2 A , E = σ ǫ correct benavides (jjb2356) – homework 05 – Turner – (59130) 2 Explanation: 004 10.0 points Two large, parallel, insulating plates are...
View Full Document

{[ snackBarMessage ]}

Page1 / 5

HW 5 - benavides(jjb2356 – homework 05 – Turner...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online