Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Please help!! I am very unsure as to how to solve this problem.... Especially th

ID: 1407905 • Letter: P

Question

Please help!! I am very unsure as to how to solve this problem.... Especially the y-component of the total electric field at P!!

Two charges, Q1= 2.20 C, and Q2= 6.90 C are located at points (0,-2.50 cm ) and (0,+2.50 cm), as shown in the figure.

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 = 2.20 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?

Explanation / Answer

a)

r = 0.025^2 + 0.065^2 = 0.00485 m

theta = tan^-1(2.5 / 6.5) = 21.04 deg

here the formula

Ex = k * q1 * cos(theta) / r^2 + k * q2 * cos(theta) / r^2

Ex = (9 * 10^9 * 2.2 * 10^-6 * * cos(21.04deg)) / 0.00485 + (9 * 10^9 * 6.9 * 10^-6 * cos(-21.04 deg) )/ 0.00485)

Ex = 1.576 * 10^7 N/C

b)

Ey = k * q1 * cos(theta) / r^2 + k * q2 * cos(theta) / r^2

Ey = (9 * 10^9 * 2.2 * 10^-6 * sin(21.04deg)) / 0.00485 + (9 * 10^9 * 6.9 * 10^-6 * sin(-21.04 deg) / 0.00485

Ey = -3.13 * 10^6 N/C

c)

magnitude = sqrt(Ex^2 + Ey^2)

magnitude = sqrt( (1.576 * 10^7)^2 + (3.13 * 10^6)^2)

magnitude = 1.6 * 10^7 N/C

d)

only the y component is non zero due to symmetry

therefore

Ey = 2 * k * q * sin(theta) / r^2

Ey = 2 * 9 * 10^9 * 2.2 * 10^-6 * sin(21.04deg) / 0.00485

Ey = 2.931 * 10^6 N/C