Three charges, q1 = +3.71nC, q2 = -1.63nC, and q3 = +8.19nC, are at the corners

Question

Three charges, q1 = +3.71nC, q2 = -1.63nC, and q3 = +8.19nC, are at the corners of an equilateral triangle, each of whose sides are of length, L, as shown in the figure below. (Figure Included Here) The angle ? is 60.0° and L = 0.410 m.

We are interested in the unmarked point midway between the charges q1 and q2 on the x axis. For starters, calculate the magnitude and direction of the electric field due only to charge q1 at this point. Down Up Left Right Submit Answer Tries 0/10

Calculate the magnitude and direction of the electric field due only to charge q2 at this point. Down Up Left Right Submit Answer Tries 0/10

Calculate the magnitude and direction of the electric field due only to charge q3 at this point. Down Up Left Right Submit Answer Tries 0/10

Now calculate the magnitude of the electric field from all three charges at a point midway between the two charges on the x axis. Submit Answer Tries 0/10

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) Submit Answer Tries 0/10

If a tiny particle with a charge q= 1.49nC were placed at this point midway between q1 and q2, what is the magnitude of the force it would feel?

Explanation / Answer

here,

q1 = 3.71 * 10^-9 C

q2 = - 1.63 * 10^-9 C

q3 = 8.19 * 10^-9 C

L = 0.41 m

electric feild due to charge q1 , E1 = k * q1 /( L/2)^2

E1 = 9 * 10^9 * 3.71*10^-9 /( 0.41/2)^2

E1 = 794.53 N/C

electric feild due to charge q2 , E2 = k * q2 /( L/2)^2

E2 = 9 * 10^9 * 1.63 *10^-9 /( 0.41/2)^2

E2 = 349.08 N/C

electric feild due to charge q3 , E3 = k * q3 /( 0.86 * L)^2

E3 = 9 * 10^9 * 8.19 * 10^-9 /( 0.86 * 0.41)^2

E3 = 584.65 N/C

as E1 and E2 are in positive x direction and q3 is in -ve y direction

E = sqrt( ( E1+E2)^2 + E3^2)

E = 1283.85 N/C

the magnitude of net electric feild is 1283.85 N/C

theta = arctan( E3/(E1+E2))

theta = arctan( 584.65 /(794.53 + 349.08))

theta = 27.08 degree

the direction of electric feild is - 27.08 degree from the x axis

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