Question

Question Completion Status: QUESTION 3 2 points Three point charges lie along a circle of radius 2.00 cm: q: -- 12.0 nC at angle 30.9,- +8.00 nC at angle 180, and , +12.0 nC at angle 330 as shown in the figure. Find the magnitude e of the electric field created by the three charges at the center of the circle. V 9: 180° 300 330 9 E = 72.0 x 10' N/C E = 32.4 x 10' N/C E = 18.0 x 10' N/C N/C E = 29.5 x 10' N/C 2 QUESTION 4 Save All Answers Close Wind PA uit tomond sumir Click Save All Answers to save all answers,

Answer 1

We know that electric field is given by:

E = kQ/R^2

Where direction of electric field will be away from positive charge and towards the negative charge.

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From above given figure, net electric field in x-direction will be: (towards right is +ve x-axis, and upward is +ve y-axis)

Ex_net = E3*cos 30 deg - E1*cos 30 deg - E2

Ex_net = k*q3*cos 30 deg/r3^2 - k*q1*cos 30 deg/r1^2 - k*q2/r2^2

r1 = r2 = r3 = r = radius of circle = 2.00 cm = 0.02 m

So,

Ex_net = (k/r^2)*(q3*cos 30 deg - q1*cos 30 deg - q2)

Ex_net = (9*10^9/0.02^2)*(12.00*10^-9*cos 30 deg - 12.00*10^-9*cos 30 deg - 8.00*10^-9)

Ex_net = -180000 N/C

From above given figure, net electric field in y-direction will be: (towards right is +ve x-axis, and upward is +ve y-axis)

Ey_net = E3*sin 30 deg - E1*sin 30 deg

Ey_net = k*q3*sin 30 deg/r3^2 - k*q1*sin 30 deg/r1^2

r1 = r2 = r3 = r = radius of circle = 2.00 cm = 0.02 m

So,

Ey_net = (k/r^2)*(q3*sin 30 deg - q1*sin 30 deg)

Ey_net = (9*10^9/0.02^2)*(12.00*10^-9*sin 30 deg - 12.00*10^-9*sin 30 deg)

Ey_net = 0 N/C

So, net electric field will be:

E_net = Ex_net i + Ey_net j

So

E_net = -180000 i + 0 j

Magnitude of net electric field will be:

|E_net| = 18.0*10^4 N/C

Correct option is C.

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