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koban
23 hours ago
15

Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance

ddd is
|F|=K|QQ′|d2|F|=K|QQ′|d2,


where K=14πϵ0K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2)ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space.


Consider two point charges located on the x axis: one charge, q1q1q_1 = -15.0 nCnC , is located at x1x1x_1 = -1.660 mm ; the second charge, q2q2q_2 = 34.5 nCnC , is at the origin (x=0.0000)(x=0.0000).


What is the net force exerted by these two charges on a third charge q3q3q_3 = 47.0 nCnC placed between q1q1q_1 and q2q2q_2 at x3x3x_3 = -1.240 mm ?


Your answer may be positive or negative, depending on the direction of the force.
Physics
1 answer:
kicyunya [3.1K]23 hours ago
5 0

Response:

Clarification:

The force between two charges, q₁ and q₂ at a distance d is represented by the formula

F = k q₁ q₂ / d²

Here, the force between charge q₁ = -15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C with distance d = (1.66 - 1.24) = 0.42 mm

k = 1 / (4π x 8.85 x 10⁻¹²)

Substituting the values into the equation

F = 1 / (4π x 8.85 x 10⁻¹²) x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 35969.4 x 10⁻³ N.

For the force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C at a distance d = (1.24 - 0) = 1.24 mm.

Substituting the values into the expression

F = 1 / (4π x 8.85 x 10⁻¹²) x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 82729.6 x 10⁻³ N

Both forces direct towards the left (away from the origin, towards the negative x-axis)

Total force = 118699 x 10⁻³

= 118.7 N.

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