Answer:
q = 3.87 x 10⁵ C
Explanation:
given,
Electric field, E = 8.60 x 10¹ = 86 N/C
radius of earth, R = 6371 Km = 6.371 x 10⁶ m
Coulomb constant, K = 9 x 10⁹ N · m²/C²
Charge on the earth = ?
the electric field at the point
inserting all the values
q = 3.87 x 10⁵ C
The electric charge on the earth is equal to 3.87 x 10⁵ C
let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Answer:
9.98 × 10⁻⁹ C
Explanation:
mass, m = 1.00 × 10⁻¹¹ kg
Velocity, v = 23.0 m/s
Length of plates D₀ = 1.80 cm = 0.018 m
Magnitude of electric field, E = 8.20 × 10⁴ N/C
drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m
density of the ink drop = 1000 kg/m^3
Now,
Time =
or
Time =
or
Time = 6.9 × 10⁻⁴ s
Now, force due to the electric field, F = q × E
where, q is the charge
Also, Force = Mass × acceleration
q × E = 1.00 × 10⁻¹¹ × a
or
a =
Now from the Newton's equation of motion
where,
d is the distance
u is the initial speed
a is the acceleration
t is the time
or
or
q = 9.98 × 10⁻⁹ C