The angle is 60.0° and L = 0.471 m . We are interested in the unmarked point mid

Question

The angle is 60.0° and L = 0.471 m. We are interested in the unmarked point midway between the charges q1 and q2 on the x axis.

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).

If a tiny particle with a charge q= 1.10nC were placed at this point midway between q1 and q2, what is the magnitude of the force it would feel?

Explanation / Answer

field due to a point charge,

E = k q / r^2

away from charge if it is positive charge and towards for negative charge.

E3 = k q3 / ( L sin(alpha))^2 (-j)

E3 = - (9 x 10^9) (9.21 x 10^-9)/ (0.471 sin60)^2 (-j)

E3 = - 498.2 N/C j

Magnitude = 498.2 N/C .......Ans

direction = downwards .......Ans

E1 + E2 = (k q1 / (L/2)^2)i + (k q2 / (L/2)^2)i

= (9 x 10^9 / (0.471 / 2)^2) (2.50 x 10^-9 + 1.40 x 10^-9)

= 632.9 N/C i

E = E1 + E2 + E3 = 632.9i - 498.2 j

magnitude = sqrt(632.9^2 + 498.2^2) =805.5 N/C .......Ans

Direction = - tan^-1( 498.2 / 632.9 ) = - 38.2 deg ......Ans

F = q E = (1.1 x 10^-9) (805.5) = 8.86 x 10^-7 N ....Ans

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