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3 years ago

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Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is

at the origin. The third charge, q3 = - 30 μC, is located at x = 2.0 m. What is the force on q2?
(a) 1.65 N in the negative x- direction
(b) 3.15 N in the positive x- direction
(c) 1.50 N in the negative x- direction
(d) 4.80 N in the positive x- direction
(e) 4.65 N in the negative x- direction.

1 answer:

victus00 [196]3 years ago

3 0

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = $\frac{-kQq}{r^2}$

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = $\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}$ = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = $\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}$ = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

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