Three positive charges are located in the x-y plane (see plot below), with Q1=2.

Question

Three positive charges are located in the x-y plane (see plot below), with Q1=2.10 C, Q2=3.20 C and Q3=4.30 C. Note that the charges are located at grid intersections, and that the x and y coordinates are in cm.

Three positive charges are located in the x-y plane (see plot below) with Q1-2.10 C, Q2-3.20pC and Q3-4.30 charges are located at grid intersections, and that the x and y coordinates are in cm Note that the 01 1 02 -1 -2 -3 -4 . 03 -5-4-3-2-1 1 2 3 x (cm) Calculate the magnitude of the electric force on Q3 due to Q2 9.51x101 N Computer's answer now shown above. You are correct Yourreceipt no. is 154-9917 Calculate the y-component of the force on Q3 due to Q1. -0.000650 N Submit Answer Incorrect. Tries 7/12 Previous Tries

Explanation / Answer

Given x and y coordinates are in cm

Q1=(-1,1)

Q2=(0,0)

Q3=(3,-2)

A)

Use Coulomb's law:

F = kqq / r²

where q and q are the charges of the two point charges in question, and r is the distance between the two. k is a proportionality constant given by:

k = 1 / 4 8.99 × 10 Nm²C²

In your case:

q = Q3 = 4.30 C = 4.30 × 10 C
q = Q2 = 3.20 C = 3.20 × 10 C

r = (0.03² + 0.02²) = 0.0013

so the magnitude of the electric force on Q3 due to Q2 is given by:

F = (8.99 × 10)(4.30 × 10)(3.20 × 10) / (0.0013)² = 95.1 N

B)

Force on Q3 due to Q1 = (8.99 × 10)(4.30 × 10)(2.10 × 10) / (0.04² + 0.03²) = 32.47 N û

where û = (1/2)[1, -1]

û is the unit vector pointing in the direction of the force! A unit vector is a vector with length = 1, that is why i have divided [1, -1] by 2.

The angle between û and the negative x-axis is 45°, so the y component of the force is;

sin(45°)(32.47) = (1/2)(32.47) N = 22.95 N

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