Three charges, q_1 = +2.63 nC, q_2 = -1.39 nC, and q_3 = +8.69 nC, are at the co

Question

Three charges, q_1 = +2.63 nC, q_2 = -1.39 nC, and q_3 = +8.69 nC, are at the corners of an equilateral triangle, each of whose sides are of length, L, as shown in the figure below. The angle a is 60.0 degree and L = 0.485 m. We are interested in the unmarked point midway between the charges q_1 and q_2 on the x axis. For starters, calculate the magnitude and direction of the electric field due only to charge q_1 at this point. Down Up Left Right Calculate the magnitude and direction of the electric field due only to charge q_2 at this point. Down Up Left Right Calculate the magnitude and direction of the electric field due only to charge q_3 at this point. Down Up Left Right Now calculate the magnitude of the electric field from all three charges at a point midway between the two charges on the x axis. Calculate the angle of the electric field relative to the positive (to the right) x-axis, with positive values up (counterclock-wise) and negative down (clockwise). (enter the answer with units of deg) If a tiny particle with a charge q = 1.50 nC were placed at this point midway between q_1 and q_2, what is the magnitude of the force it would feel?

Explanation / Answer

The electric field of a point charge is given by:

E k q/r^2

E1 = k q1/r1^2

E1 = 9 x 10^9 x 2.63 x 10^-9/0.485^2 = 100.63 N/C

Hence, E1 = 100.63 N/C

Diretion- Right.

E2 = k q2/r2^2

The Elecxtric field due to q2 will be zero since, r2 = 0.

E3 = k q3/r3^2

E3 = 9 x 10^9 x 1.39 x 10^-9/0.485^2 = 332.5 N/C

Hence, E3 = 332.5 N/C

Direction- Down

At the mid point,

r1 = r2 = 0.485/2 = 0.243 m

r3 = sqrt (0.485^2 - 0.243^2) = 0.419 m

E1 = k q1/r1^2 = 9 x 10^9 x 2.63 x 10^-9/0.243^2 = 400.85 N/C

E2 = k q2/r2^2 = 9 x 10^9 x 1.39 x 10^-9/0.243^2 = 211.85 N/C

E3 = k q3/r3 = 9 x 10^9 x 8.69 x 10^-9/0.419^2 = 445.47 N/C

Ex = E1 - E2 = 400.85 - 211.85 = 189 N/C

Ey = E3 = 445.47 N/C

E = sqrt (Ex^2 + Ey^2) = sqrt (189^2 + 445.47^2) = 483.91 N/C

Hence, E = 483.91 N/C

theta = tan^-1(445.47/189) = 67 deg

Hence, theta = 67 deg

force will be:

F = q E = 1.5 x 10^-19 x 483.91 = 7.26 x 10^-17 N

Hence, F = 7.26 x 10^-17 N

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