. Use Coulomb\'s law to determine themagnitude and direction of the electric fie

Question

. Use Coulomb's law to determine themagnitude and direction of the electric field atpoints A and B in Fig. 16-57due to the two positive charges(Q = 5.0 µC) shown. Are your resultsconsistent with Fig. 16-31b?

1.a.What is the magnitude at point A_____N/C

   b. direction in degree________ (above or belowhorizontal)

2.a.What is the magnitude at pointB_____N/C

   b. direction in degree________ (above or belowhorizontal)

      .B    .A

   .+Q             .+Q

     ____5.0cm__/______5.0cm_____/________10.0cm______

Thedistance from the first charge is to point B is 5.0 cm and fromPoint B to point A is 5.0cm and from Point A to the second chargeis 10.0 cm. The perpendicular distance from the charge to point Ais 5.0cm.

P.SWhen setting up the diagram to calculate the x nad y vallues pleaseexplain the direction and the angles formed.

Explanation / Answer

Point A In the x direction r1=sqrt(25+100)/100 m r2=sqrt(25+100)/100 m at Point A the fields in the x cancel. Let's look at y C*Q*2*(100^2/125)/sqrt(2) upward 711.8*10^4 N/C in the positive y (90 degrees above the horizon) Point B in the x direction r1=5*sqrt(2)/100 m r2=sqrt(25+15^2)/100 m from the left C*Q*(1/r1^2)/sqrt(2) from the right -C*Q*(1/r2^2)*0.15/r2 add them together Ex=C*Q*((1/r1^2)/sqrt(2)- (1/r2^2)*0.15/r2) 6509847 N/C in the plus x direction now the y from the left C*Q*(1/r1^2)/sqrt(2) from the right C*Q*(1/r2^2)*0.05/r2 add them together Ey=C*Q*((1/r1^2)/sqrt(2)+ (1/r2^2)*0.05/r2) 9693014.1 upward The resultant is sqrt(Ex^2+Ey^2) 11676156.5 N/C direction is atan(Ey/Ex) 56.11 degrees above the horizontal ________________________-

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.