{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# OHW 2 - terry(ect328 oldhomework 02 Turner(59130 This...

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

terry (ect328) – oldhomework 02 – Turner – (59130) 1 This print-out should have 13 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 Three charges are arranged as shown in the figure. The Coulomb cosnstant is 8 . 98755 × 10 9 N · m 2 / C 2 . x y 3 . 7 nC 4 . 3 m 0 . 9 nC 0 . 6 nC 1 . 9 m Find the magnitude of the electrostatic force on the charge at the origin. Correct answer: 2 . 10413 nN. Explanation: Let : ( x 0 , y 0 ) = (0 m , 0 m) , q 0 = 0 . 9 nC , ( x 1 , y 1 ) = (4 . 3 m , 0 m) , q 1 = 3 . 7 nC , ( x 2 , y 2 ) = (0 m , 1 . 9 m) , and q 2 = 0 . 6 nC . x y q 1 x 1 q 0 q 2 y 2 F θ The force vector F 10 (in the hatwider 10 direction, ˆ ı ) be- tween charge 9 × 10 10 C and 3 . 7 × 10 9 C is vector F 10 = + k e q 0 q 1 x 2 1 ( ˆ ı ) = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 0 . 9 nC) (3 . 7 nC) (4 . 3 m) 2 ˆ ı = (1 . 61863 × 10 9 N)ˆ ı . The plus sign means the force vector F 10 is towards the positive x -axis. The force vector F 20 (in the hatwider 20 direction, +ˆ ) between charge 9 × 10 10 C and 6 × 10 10 C is vector F 20 = + k e q 0 q 2 y 2 2 (+ˆ ) = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 0 . 9 nC) (0 . 6 nC) ( 1 . 9 m) 2 ˆ = ( 1 . 3444 × 10 9 N)ˆ  . The minus sign means the force vector F 20 is towards the negative y -axis. The magni- tude of the total force exerted on the charge 9 × 10 10 C is equal to bardbl vector F bardbl = radicalBig F 2 10 + F 2 20 = bracketleftBig (1 . 61863 × 10 9 N) 2 + ( 1 . 3444 × 10 9 N) 2 bracketrightBig 1 2 = 2 . 10413 × 10 9 N = 2 . 10413 N . 002 (part 2 of 2) 10.0 points What is the angle θ between the electrostatic force on the charge at the origin and the pos- itive x -axis? Answer in degrees as an angle between 180 and 180 measured from the positive x -axis, with counterclockwise posi- tive. Correct answer: 39 . 7122 . Explanation: The angle between vector F and the negative y - axis is θ = arctan parenleftbigg 1 . 3444 × 10 9 N 1 . 61863 × 10 9 N parenrightbigg = 39 . 7122 .

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

View Full Document
terry (ect328) – oldhomework 02 – Turner – (59130) 2 keywords: 003 (part 1 of 2) 10.0 points A charge Q is distributed uniformly along the x axis from x 1 to x 2 . The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 Which of the following integrals is correct for the magnitude of the electric field at x 0 on the x axis? Assume that x 0 > x 2 > x 1 and k e = 1 4 πǫ 0 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}