Apositively charged bead having a mass of 1.00 g falls from rest in a vacuum from a height of 5.00 m in a uniform vertical electric field with a magnitude of 1.00 ✕ 104 n/c. the bead hits the ground at a speed of 21.9 m/s.
a) determine the direction of the electric field (up or down)
b)determine the charge on the bead âµc

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

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//