A positive charge with a mass of 3 kg is placed on a string and a negative charge of the same amount (magnitude) but negative is placed directly to the right of this charge when in equilibrium. Also, the angle with the vertical of the string is 43.24 degrees when the charge is in equilibrium. Then the negative charge is moved directly below the charge on the string and it is placed twice as far as it originally was to the charge on the string. What is the amount of tension in the string now in Newtons?
Fyafter = T2 - F2e - m
g = 0
T2 = F2e + m g
and the electric force is F2e = k q2 / r22
T2 = k q2 / r22 + m g
given r2 = 2 r1
Fx
before = F1e - T1 x = 0
T1x = T1 sin
F1e - T1 x = 0
F1e - T1 sin =
0
Fybefore = T1 y - m g = 0
T1y = T1 cos
T1 cos - m g = 0
T1 = m g / cos
k q2 / r12 - ( m g /
cos ) sin
=
0
k q2 / r12 = m g tan
k q2 / (r22 / 2 )2 =
m g tan
k q2 / r22 = m g tan /
4
T2 = m g tan / 4 + m g
= mg ( tan / 4 + 1 )
= 3 x 9.8 x ( ( tan 43.24 /4 ) + 1 )
T2 = 29.4 x 1.253
T2 = 36.31177 N
the amount of tension in the string is T2 = 36.31177 N
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