E-field 2-solutions

E-field 2-solutions - gilvin (jg47854) - 4. E-field 2 -...

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

View Full Document Right Arrow Icon
gilvin (jg47854) - 4. E-field 2 - meyers - (21235) This print-out should have 23 questions. The electric field E 1 is Multiple-choice questions may continue on 1 q 1 the next column or page - find all choices E 1 = 2 before answering. This homework is due Friday, February 4, at midnight Tucson time. 4 π ϵ 0 ( ) 2 sin θ d r 2 + 2 d 001 (part 1 of 4) 10.0 points Since sin θ = [ 2 ( ) 2 ] 1/2 , A good model for a radio antenna is a dipole. In the figure below P 1 is on the perpendic- ular bisector of the dipole, and P 2 is along the axis of the dipole in the direction of the dipole E 1 = r 2 + 1 [ 4 π ϵ 0 d 2 q d ( ) 2 ] 3/2 vector p. P 1 and P 2 are both a distance 100 m from the center of the dipole. The magnitude of each of the charges is 4.3 µC . r 2 + 1 q d d 2 P 1 (0,100 m) 4 π ϵ 0 r 3 , for r d , so 1 q d E 1 ≈ − (−100 m, 0) ≈ − + − 4 π ϵ 0 1 4 π ϵ 0 p r 3 (4.3 × 10 −6 C)(1 m) p (100 m) 3 P 2 1 m p What is the magnitude of the electric field at P 1 ? Hint: The direction of the dipole vector p is from the negative charge to the positive charge. Correct answer: 0.0386465 N/C. Explanation: Let : r = 100 m , d = 1 m , and q = 4.3 µC = 4.3 × 10 −6 C . P 1 E + E ≈ −0.0386465 N/C p |E 1 | ≈ 0.0386465 N/C . The cos θ components cancel due to symme- try. 002 (part 2 of 4) 10.0 points What is the direction of the electric field E 1 at P 1 ? 1. In the direction of r. 2. In a direction perpendicular to both the dipole vector p and r. 3. In the direction opposite to that of the dipole vector; e.g., −p. correct 4. In the direction of the dipole vector; e.g., p. 5. In a direction perpendicular to the dipole vector; e.g., p. Explanation: The electric field goes from positive charge P 2 + − to negative charge. p The dipole vector p is shown in the figure in the question.
Background image of page 1

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

View Full DocumentRight Arrow Icon
gilvin (jg47854) - 4. E-field 2 - meyers - (21235) 2 4. In a direction perpendicular to the dipole 003 (part 3 of 4) 10.0 points What is the magnitude of the electric field at P 2 ? Correct answer: 0.0772929 N/C. Explanation: i i i i vector; e.g., p. 5. In the direction of the dipole vector; e.g., p. correct Explanation: The electric field goes from positive charge to negative charge. 005 (part 1 of 3) 10.0 points E 2 = q 4 π ϵ 0 i i i i 1 ( ) 2 r − d 2 ( ) 2 1 i Two point-charges at fixed locations pro- ( ) 2 i i duce an electric field as shown below. r + d 2 i ( ) 2 = q 4 π ϵ 0 i i i i i r + d r − d i 2 2 i ( ) 2 ( ) 2 i r − d 2 h ( ) h r + d 2 A B q 4 π ϵ 0 i i i 4 r d 2 i i for r d , r 4 i , Y so X E 2 1 2 π ϵ 0 1 2 π ϵ 0 1 2 π ϵ 0 1 2 π ϵ 0 q d r 3 q d p r 3 (4.3 × 10 −6 C)(1 m) p (100 m) 3 (4.3 × 10 −6 C)(1 m) p (100 m) 3 A negative charge placed at point Y would move 1. along an equipotential plane. 2. toward charge B. correct
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

Page1 / 9

E-field 2-solutions - gilvin (jg47854) - 4. E-field 2 -...

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