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.

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

the part of the wave he is travelling on is the crest