Respuesta :
Answer:
a)
down direction.
b)
3.82 µC
Explanation:
a)
Consider the motion of the positively charged bead in vertical direction
y = vertical displacement of charged bead = 5 m
a = acceleration of charged bead = ?
v₀ = initial velocity of bead = 0 m/s
v = final velocity of bead = 21.9 m/s
using the equation
v² = v₀² + 2 a y
inserting the values
21.9² = 0² + 2 a (5)
a = 47.96 m/s²
m = mass of the bead = 1 g = 0.001 kg
F = force by the electric field
Force equation for the motion of the bead in electric field is given as
mg + F = ma
(0.001) (9.8) + F = (0.001) (47.96)
F = 0.0382 N
Since the electric force due to electric field comes out to be positive, the electric force acts in down direction. we also know that a positive charge experience electric force in the same direction as electric field. hence the electric field is in down direction.
b)
q = magnitude of charge on the bead
E = electric field = 1 x 10⁴ N/C
Electric force is given as
F = q E
0.0382 = q (1 x 10⁴)
q = 3.82 x 10⁻⁶ C
q = 3.82 µC
