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?
5.79×106 N/C

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 = 3.30 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= 3.30 C, and Q2= 5.00 C are located at points (0,-3.00 cm ) and (0,+3.00 cm), as shown in the figure.

Explanation / Answer

distance of point P from q1 & q2 = r1 = r2 = sqrt(x^2+y^2)

r1 = r2 = sqrt(6.5^2+3^2) = 7.2 cm = 0.072 m

tantheta = y/x = 3.3/6.5

theta = 26.92 degrees

E1x = E1*cos27 = k*q1*cos27/r1^2 = (9*10^9*3.3*10^-
6*cos27)/0.0716^2 = 5.16*10^6 N/C

E1y = E1*sin27 = k*q1*sin27/r1^2 = (9*10^9*3.3*10^-
6*sin27)/0.0716^2 = 2.63*10^6 N/C

E2x = E2*cos27 = k*q2*cos27/r2^2 = (9*10^9*5*10^-
6*cos27)/0.0716^2 = 7.82*10^6 N/C

E2y = -E28sin27 = k*q2*sin27/r2^2 = (9*10^9*5*10^-

6*sin27)/0.0716^2 = -3.98*10^6 N/C

x-component of the total electric field at P

Ex = E1x + E2x = 5.16*10^6 + 7.82*10^6 = 12.98*10^6 N/C

y-component of the total electric field at P

Ey = E1y + E2y = 2.63*10^6 -3.98*10^6 = -1.35*10^6 N/C

magnitude of the total electric field at P

E = sqrt(Ex^2 + Ey^2) = 13.05*10^6 N/C

E1 = E2

E1y = -E2y

E1x = E2x

Ex = E1 + E2x = 2*E*c0s27/r^2 = (2*9.*10^9*3.3*10^-6*cos27)/0.0716^2

Ex = 10.32*10^6 N/C

Ey = E1y + E2y = 0

E = Ex + Ey = 10.32*10^6

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Force F = E*q = 10.32*10^6*1.6*10^-19 = 1.65*10^-12 N <<<<------answer

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