Question

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.

Answer 1

3.15 N towards the positive x-axis

Step-by-step 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 =
`(-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 =
`(-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 =
`(-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 = 3.15 N towards the positive x-axis