The tiny ball at the end of the thread shown in fig. 24-8 has a mass of 0.60g and is in a horizontal electric field of strength 700N/C. It is in equilibrium in the position shown. What are the magnitude and sign of the charge on the ball?
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Answer:
−3.1×10∧-6
Explanation:
Given:
m=0.6×10-3
kg
E=700 N/C
α=20°
g=9.8m/s2
Solution:
Consider the forces which acting on the tiny ball q.
Newton's Second Law:
F = ma (1)
Of (1)
⇒T + m g + f = ma (2),
where T is tension force,
m g is gravity
f
is force of electric field
Projections of the vectors:
OX: −T sin α + f = 0 (3)
OY: T cos α − mg = 0 (4)
Force of electric field:
f = E q (5)
(5) in (3): T sin α = E q (6)
Of (4)
T cos α = mg (7)
We divide (6) on (7) term by term:
tan α =
E q
mg
(8)
Of (8)
q =
mg tan α
E
(9)
Of (9)
q = 3.1 × 10−6C
From Figure
sign of the charge: q=-3.1×10-6 C
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