Physics, 26.08.2020 17:01 AngelOfLove

Two charged particles, q1 and q2, are located on the x-axis, with q1 at the origin and q2 initially atx1 = 14.7 mm. In this configuration, q1 exerts a repulsive force of 2.62 µN on q2. Particle q2 is then moved tox2 = 18.0 mm. What is the force (magnitude and direction) that q2 exerts on q1 at this new location? (Give the magnitude in µN.)

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Answer from: pierceybbpow8o39

To magnitude of the force that q_2 exerts on q_1 at this new location is 1.75 µN.

Given the following data:

Initial distance of q_2

= 14.7 mm to m = 0.0147 meter.Repulsive force = 2.62 µNFinal distance of q_2

= 18.0 mm to m = 0.018 meter.

Scientific data:

Coulomb's constant = $8.99 \times 10^9\; Nm^2/C^2$

To calculate the force (magnitude and direction) that q_2 exerts on q_1 at this new location, we would apply Coulomb's law:

The force in an electric field.

Mathematically, the force in an electric field is given by this formula:

$F=\frac{kq_1q_2}{r^2}$

Where:

k is Coulomb's constant.q is the charge.r is the distance.

$q_1q_2 =\frac{F}{k} \times r^2\\\\q_1q_2 =\frac{2.62 \times 10^{-6}}{8.99 \times 10^9} \times 0.0147^2\\\\q_1q_2 = 2.914 \times 10^{-16} \times 0.00021609\\\\q_1q_2 = 6.30 \times 10^{-20}\;C^2$

For the force, when r = 0.018:

$F =\frac{8.99 \times 10^9 \times 6.30 \times 10^{-20}}{0.018^2} \\\\F =\frac{5.66 \times 10^{-10}}{0.000324} \\\\F = 1.75 \times 10^{-6}$

Note: 1 µN = $1 \times 10^{-6}\;N$

Force, F = 1.75 µN.

In conclusion, the direction of the force at the new location is outward from the origin along X_2 .

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Two charged particles, q1 and q2, are located on the x-axis, with q1 at the origin and q2 initially...

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