{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 9 - terry(ect328 homework 09 Turner(59130 This print-out...

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

terry (ect328) – homework 09 – Turner – (59130) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Three identical point charges, each of mass 150 g and charge + q , hang from three strings, as in the figure. The acceleration of gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 9 . 8 m / s 2 10 cm 150 g + q 10 cm 150 g + q 45 150 g + q If the lengths of the left and right strings are each 10 cm, and each forms an angle of 45 with the vertical, determine the value of q . Correct answer: 0 . 80885 μ C. Explanation: Let : θ = 45 , m = 150 g = 0 . 15 kg , L = 10 cm = 0 . 1 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . Newton’s 2nd law: summationdisplay F = m a . Electrostatic force between point charges q 1 and q 2 separated by a distance r F = k e q 1 q 2 r 2 . All three charges are in equilibrium, so for each holds summationdisplay F = 0 . Consider the forces acting on the charge on the right. There must be an electrostatic force F acting on this charge, keeping it balanced against the force of gravity m g . The electro- static force is due to the other two charges and is therefore horizontal. In the x -direction F T sin θ = 0 . In the y -direction T cos θ m g = 0 . These can be rewritten as F = T sin θ and m g = T cos θ . Dividing the former by the latter, we find F m g = tan θ , or F = m g tan θ (1) = (0 . 15 kg) (9 . 8 m / s 2 ) tan 45 = 1 . 47 N . The distance between the right charge and the middle charge is L sin θ , and the distance to the left one is twice that. Since all charges are of the same sign, both forces on the right charge are repulsive (pointing to the right). We can add the magnitudes F = k e q q ( L sin θ ) 2 + k e q q (2 L sin θ ) 2 or F = 5 k e q 2 4 L 2 sin 2 θ . (2) We have already found F , and the other quan- tities are given, so we solve for the squared charge q 2 q 2 = 4 F L 2 sin 2 θ 5 k e (3)

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

View Full Document
terry (ect328) – homework 09 – Turner – (59130) 2 or, after taking the square root (we know q > 0) and substituting F from Eq. 1 into Eq. 3 and solving for q , we have q = 2 L sin θ radicalbigg m g tan θ 5 k e = 2 (0 . 1 m) sin 45 × radicalBigg (0 . 15 kg) (9 . 8 m / s 2 ) tan 45 5 (8 . 98755 × 10 9 N · m 2 / C 2 ) = 8 . 0885 × 10 7 C = 0 . 80885 μ C . 002 10.0 points An electron moves at 3 . 5 × 10 6 m / s into a uniform electric field of magnitude 1194 N / C. The charge on an electron is 1 . 60218 × 10 19 C and the mass of an electron is 9 . 10939 × 10 31 kg . The field is parallel to the electron’s velocity and acts to decelerate the electron. How far does the electron travel before it is brought to rest? Correct answer: 2 . 91662 cm. Explanation: Let : v = 3 . 5 × 10 6 m / s , q e = 1 . 60218 × 10 19 C , m = 9 . 10939 × 10 31 kg , and E = 1194 N / C . The kinetic energy K = 1 2 m v 2 is depleted by the amount of work done by the electric force F = q e E on the particle: W = integraldisplay F dx = F x = q e E x since the force is constant. When the electron comes to rest, all its kinetic energy has been converted, so 1 2 m v 2 = q e E x .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

HW 9 - terry(ect328 homework 09 Turner(59130 This print-out...

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

View Full Document
Ask a homework question - tutors are online