Three charges, q1 = +2.91nC, q2 = -1.38nC, and q3 = +9.37nC, are at the corners

Question

Three charges, q1 = +2.91nC, q2 = -1.38nC, and q3 = +9.37nC, are at the corners of an equilateral triangle, each of whose sides are of length, L, as shown in the figure below.

The angle is 60.0° and L = 0.529 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.

Calculate the magnitude and direction of the electric field due only to charge q2 at this point.

Calculate the magnitude and direction of the electric field due only to charge q3 at this point.

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= 1.41nC were placed at this point midway between q1 and q2, what is the magnitude of the force it would feel?

Explanation / Answer

.           q3

. q1                          x                          q2   

We used the electric field equation

E = k q/ r2         

. q1= 2.91 10-9 C

. x = L = 0.529 m            r= L/2

E1 = 9 109   2.91 10-9 / 0.26452      E1= 374.36 N/C    in direction x axis positive (right)

. q2= -1.38 10-9 C

E2 = 9 109    (-1.38 10-9)/ 0.26452     E2 = -177.53 N/C    in direction x axis positive (right)

. q3 = 9.37 10-9 C

We are looking for the height (y) using trigonometry

.   L2 = (L/2)2 + y2         y2 = L2 – L2/4 = ¾ L2               y = sqrt(3)   L/2 = sqrt(3) 0.529/2    

.   y= 0.458 m

E3 = 9 109   9.37 10-9 / 0.4582         E3= 401.85 N/C in direction y axis negative (down)

The electric field is a vector magnitude

Et = E1 +E2 + E3      

Et = Etx i + Ety   j      Et = (E1 +E2)   i + E3 j

Et = (374.36+ 177.53) i   - 401.85 j

Et = (551.89 i - 401.85 j )   N/C

Magnitude

Et2 = 551.892 + 401.852        Et= sqrt(466065) = 682.69 N/C

Tg angle = -401.85/551.89      angle = Tg-1( -0.728) = -36°

Angle 36° below f the axis x

Angle measured counterclockwise 360-36 = 324°

Measuring angles in a clockwise direction 36°

Last Part

. qo= 1.41 10-9 C

Ft = (F1+ F2) i + F3 j                 Ft = ( E1 qo +E2 qo) i + E3 qo j    Ft = qo Et

Ft = 1.41 10-9 682.69   Ft = 962.59 10-9 N = 9.62 10-7 N    angle = -36°

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