How do scientist know that seismic waves can be either compressional or transverse
1 answer:
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Answer:
- Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
- Uncertainty in momentum (∆P) = ?
- Planck's constant (h) = 6.26 × 10⁻³⁴ Js
The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
Given values:
Length of non-conducting rod, l = 1.20 m
Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C
Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C
Distance from point P of each rod, x = 60 cm = 0.60 m
Calculation of Net electric force exerted on point P:
Consider an electron released at point P, then the net electric force exerted will be given as:
F = e. E_net - ( 1 )
Step 1:
The net electric field value is given as:
E_net = E₁ cos Φ + E₂ cos Φ
= 2E₁ cos Φ -( 2 )
where, E₁ & E₂ are electric fields due to positive and negative rod
respectively.
Φ is phase angle
Step 2:
The electric field due to positive rod is given as:
E₁ = k (λ/r) - ( 3 )
where, k is Coulomb's force constant
λ is linear charge density
r is distance between point P and half of the rod.
Now, the linear charge density is given as:
λ = Charge/length = Q/x
The value of r is given as:
r = √x²-a²
where, x is length of rod
a is half length of rod
Applying values in above equation, we get:
r = √x²-(x/2)²
r = √(1.20 m)²-(1.20/2)²
= √1.08
= 1.04 m
Substituting all the determined values in equation 3 we get:
E₁ = k (λ/r)
= k [(Q/x)/r]
= k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 3:
Similarly, the electric field due to negative rod is given as:
E₂ = k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 4:
Consider equation 2:
E_net = 2E₁ cos Φ
From the figure we get the phase angle as:
Φ = tan⁻¹ (0.60 m/0.60 m)
= tan⁻¹ ( 1 )
= π/4
Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:
E_net = 2(1.803×10⁴ N/C) cos π/4
= 2(1.803×10⁴ N/C) (0.5)
= 18030 N/C
Step 5:
Consider equation 1 :
F = e. E_net
where, e is charge on an electron
Applying values in above equation we get:
F = (1.6 × 10⁻¹⁹ C)(18030 N/C)
= 2.885 × 10⁻¹⁵ N
Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
Learn more about electric force here:
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