Two charges, Q1- 3.30 HC, and Q2- 7.00 uC are located at points (0,-3.50 cm) and

Question

Two charges, Q1- 3.30 HC, and Q2- 7.00 uC are located at points (0,-3.50 cm) and (0,+3.50 2 cm), as shown in the figure. What is the magnitude of the electric field at point P, located at (4.50 cm, 0), due to Q1 alone? 9.13x106 N/C You are correct. Previous Tries What is the x-component of the total electric field at P? 1.08 1049 N/C By the principle of linear superposition, the total electric field at position P is the vector sum of the electric field contribution from charge Q1 and Q2. Submit Answer Incorrect. Tries 2/20 Previous Tries What is the y-component of the total electric field at P? Submit Answer Tries 0/20 What is the magnitude of the total electric field at P? Submit Answer Tries 0/20 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? Submit Answer Tries 0/20 Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P? Submit Answer Tries 0/20

Explanation / Answer

at point P,

angle made by Q2 is tan theta = Y/x

tan theta = 3.5/4.5

theta1 = 37.87 deg

angle made by Q1 is tan theta2 = -3.5/4.5

theta 2 = -37.87 deg

X component of electric field Ex du to Q1 is Ex = E Cos (-37.87)

Ex due to Q1 = 9*10^9*3.3*10^-6*cos(-37.87)/(0.045^2 + 0.035^2)

Ex due to Q1 = 7.213 *10^6 N/C

X component of electric field Ex du to Q2 is Ex = E Cos (37.87)

Ex due to Q2 = 9*10^9* 7*10^-6*cos(-37.87)/(0.045^2 + 0.035^2)

Ex due to Q2 = 15.3 *10^6 N/C
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total X component of EF = 7.213 + 15.3 = 22.51 *10^6 N/C

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due to Q1, Y component is Ey = E sin theta

Ey = KQ1/r^2 * sin theta

Ey = (9*10^9* 3.3*10^-6* sin -37.87/(0.035^2 + 0.045^2)

Ey due to Q1 = -5.61 *10^6 N/C

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Ey due to Q2 = (9*10^9* 7*10^-6* sin 33.7)/(0.035^2 + 0.045^2)

Ey = 10.755 *10^6 N/C

--------------------------------

total Y component of Electirc field

Ey = -5.61 + 10.755 = 5.145 *10^6 N/C

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total Eelcric field E^2 = Ex^2 + Ey^2

E^2 = (22.61^2 + 10.755^2)

E = 25*10^6 N/C
-----------------------------------

ENnet at P ue same charges is only due to x comp

as Y componets in opposite direction gets cancelled

so

Enet = 2Ex cos theta

Enet = 2* 9*10^9 *3.3*10^-6 * cos (37.87)/(0.035^2 +0.045^2)

Enet = 14.42 *10^6 N/C

------------------------------------

Electric force F = Eq

F = 14.42*10^6* 1.6*10^-19

F = 2.307 *10^-12 N

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