Q = magnitude of charge on each of the two point charge = 3.60 mC = 3.60 x 10⁻³ C
r = distance between the two point charges = 9.3 cm = 0.093 m
k = constant = 9 x 10⁹ Nm²/C²
F = magnitude of the force between the two point charges = ?
according to coulomb's law , force between two charges is given as
F = k Q²/r²
inserting the values
F = (9 x 10⁹) (3.60 x 10⁻³)²/(0.093)²
F = 1.35 x 10⁷ N
Answer:
The field lines go out of Earth near Antarctica, enter Earth in northern Canada, and are not aligned with the geographic poles.
Explanation:
Sitting = no movement
KE=0
Assuming the two functions are sine functions.
From the problem:
T = 1.5
So
b = 2π/T
b = 2π/1.5
b = 4π/3
The first sine function has an equation
y1 = sin 4π/3x
and since they have a difference in phase of π/6, the other sine function has an equation
y2 = sin 4π/3x + π/6
Their difference is simply π/6.
If we substitute 0.5 s to each equation,
y1 = 2π/3
y2 = 5π/6
Answer:
3.528×10⁻⁸ V.
Explanation:
Electric Potential: This can be defined as the work done in an electric field in moving a unit charge from infinity to any point. The S.I unit of electric potential is Volt (V) or J/C.
From the question,
V = Fd................ Equation 1
Where V = Electric Potential, F = force experienced by the charge, d = distance.
Given: F = 3.6×10⁻⁴ N, d = 9.8×10⁻⁵ m.
Substitute into equation 1
V = 3.6×10⁻⁴( 9.8×10⁻⁵)
V = 3.528×10⁻⁸ V.