What is the magnitude of the electric field at point P, located at (6.50 cm, 0),

Question

What is the magnitude of the electric field at point P, located at (6.50 cm, 0), due to Q1 alone?

What is the x-component of the total electric field at P?

What is the y-component of the total electric field at P?

What is the magnitude of the total electric field at P?

Now let Q2 = Q1 = 3.80 ?C. Note that the problem now has a symmetry that you should exploit in your solution. What is the magnitude of the total electric field at P?

Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

Two charges, Q1= 3.80 ?C, and Q2= 5.00 ?C are located at points (0,-3.50 cm ) and (0,+3.50 cm), as shown in the figure. 2 2

Explanation / Answer

Given that

Two charges, Q1= 3.80 C, and Q2= 5.00 C

Now the distance between the charge Q1 and the point P is

r1 =(3.502 +6.502)1/2 =7.382cm

and the distance between the charge Q2 and the point P is

r2 =(3.502 +6.502)1/2 =7.382cm

Now the direction is given by

tantheta =3.50/6.50==>theta =28.30degrees

The electric field due to the charge Q1 at the point P is given by

E1 =KQ1/r12 =(9*109)(3.80*10-6C)/(7.382*10-2)2=0.627*107N/C =6.27*106N/C

and the electric field due to charge Q2 at the point P is given by

E2 =KQ2/r22 =(9*109)(5.00*10-6C)/(7.382*10-2)2=0.627*107N/C =8.25*106N/C

Therefore the x-component of the total electric field at point P is

Ex =E1costheta+E2costheta

=6.27*106N/C*cos(28.30)+8.25*106N/C*cos(28.30)=5.520*106N/C+7.263*106N/C

=1.2783*107N/C

And the y the component of the total electric field at point P is

Ey =-E1sintheta+E2sintehta

=-6.27*106N/C*sin(28.30)+8.25*106N/C*sin(28.30)=-2.972*106N/C+3.911*106N/C

=0.0939*107N/C =9.39*105N/C

The magnitude of the total electric field at point P is

E =(Ex2+Ey2)1/2=((1.2783*107)2 +(0.0939*107)2)1/2

   =1.281*107N/C

Now let Q2 = Q1 = 3.80 C =3.80*10-6C

The total electric field at point P is

Ex =2E1costheta

Ey =0

Total electric field is E=2E1costheta =2*6.27*106N/C*cos(28.30)=1.104*107N/C

Now the forcec on the electron placed at point P is

F =QE =1.104*107N/C*1.6*10-19C =1.7664*10-12N

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