Two charges, Q1= 2.00 C, and Q2= 5.50 C are located at points (0,-2.00 cm ) and

Question

Two charges, Q1= 2.00 C, and Q2= 5.50 C are located at points (0,-2.00 cm ) and (0,+2.00 cm), as shown in the figure.

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

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

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

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

E) Now let Q2 = Q1 = 2.00 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?

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

Explanation / Answer

A) due to Q1,the distance of P from Q1 is sqrt(6.5^2+2^2) = 6.8 cm = 0.068 m

E1 = k*Q1/r^2 = (9*10^9*2*10^-6)/(0.068^2) = 3.9 *10^6 N/C

B) theta1 is the angle between the x-axis and E1

then theta1 = tan^(-1)(2/6.5) = 17.1 degree

Ex1 = E1*cos(theta1) = 3.9*10^6*cos(17.1) = 3.72*10^6 N/C

Ex2 = E2*cos(17.1) = (9*10^9*5.5*10^-6*cos(17.1))/(0.068^2) = 10.23*10^6 N/C

Ex = (3.72+10.23)*10^6 = 13.95*10^6 N/C

C) Ey1 = E1*sin(17.1) = 3.9*10^6*sin(17.1) = 1.14*10^6 N/C

Ey2 = E2*sin(17.1) = (9*10^9*5.5*10^-6*sin(17.1))/(0.068^2) = 3.14*10^6 N/C

Ey = Ey1-Ey2 = (1.14-3.14)*10^6 = -2*10^-6 N/C

D) E = sqrt(Ex^2+Ey^2) = sqrt(13.95^2+2^2)*10^6 = 14.1*10^6 N/C

E) Ex = 2*Ex1 = 7.44*10^6 N/C

Ey = 0 N/C

E = sqrt(Ex^2+Ey^2) = Ex = 7.44*10^6 N/C

F) Force is F = q*E = 1.6*10^-19*7.44*10^6 = 1.19*10^-12 N

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