Two charges, Q_1= 2.20 mu C, and Q_2=6.10 mu c, arc located at points (0,-250 cm

Question

Two charges, Q_1= 2.20 mu C, and Q_2=6.10 mu c, arc located at points (0,-250 cm) and (0,+250 cm), as shown in the figure. What is the magnitude of the electric field at point P, located at (6.00 cm,0), due to Q_1 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? addition and recall that Ep = squareroot( Epx^2 + Epy^2). What is the angle (with respect to the x-axis) of the total electric field / at point P (use deg as the symbol for degrees)?

Explanation / Answer

Electric feild (E) = kQ/r^2 // k=1/(4*pi*epsilon) = 9*10^9

Displacement from P to Q1 = sqrt(2.5^2+ 6^2) cm = 6.5cm

sin(theta) =2.5/6.5

theta = 22.6 degrees

Electric feild at point p due to Q1 = 9*10^9 x 2.20*10^-6/.065^2

E1 = 4.686*10^6 N/C

E1 = E1cos(theta) i^ + E1sin(theta) j^

= 4.326*10^6 N/C i^ + 1.8*10^6 N/C j^

Similarly,

Electric field at point P due to E2 = 9*10^9 x 6.10 * 10^-6 / .065^2

E2 = 12.99*10^6 N/C

theta = -22.6 degrees

so E2 = 11.99*10^6 N/C i^ - 4.99*10^6 N/C j^

Total elelectric feild E = E1 + E2 = 16.316*10^6 N/C i^ - 3.19*10^6 N/C j^

Now, X component = 16.316*10^6 N/C
Y component = -3.19*10^6 N/C

Total Electric field= sqrt(16.316^2+ -3.19^2)*10^6 N/C = 16.62*10^6 N/C

theta = tan-1(-3.19/16.16) = -11.062 deg

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