When body of mass 1.00 kg is suspended from a certain of light string?
Answers
Where
→
T
= Tension Force
→
F
g= Gravitational Force
→
F
e= Electric Force
→
T
=
Tension Force
→
F
g
=
Gravitational Force
→
F
e
=
Electric Force
Such that
→
F
g=mg
→
F
e=q
→
E
The charged cork ball is in equilibrium. Thus,
∑
→
F
x=0 ∑
→
F
y=0
For the forces along the x-axis:
∑
→
F
x=
→
F
ex+(−
→
T
x)=0then,
→
F
ex=
→
T
xSince
→
T
x=
→
T
sinθ,
→
F
ex=
→
T
sinθ
→
T
=
→
F
ex
sinθ
(1)
For the forces along the y-axis:
∑
→
F
y=
→
T
y+
→
F
ey+(−vecFg)=0then,
→
T
y+
→
F
ey=
→
F
gSince
→
T
y=
→
T
cosθ, and
→
F
g=mg
→
T
cosθ+
→
F
ey=mg (2)
Substituting (1) to (2):
(
→
F
ex
sinθ
)cosθ+
→
F
ey =mg
→
F
excotθ+
→
F
ey =mg but
→
F
ex=q
→
E
x
→
F
ey=q
→
E
y q
→
E
xcotθ+q
→
E
y =mg
isolating q gives us
q=
mg
→
E
xcotθ+
→
E
y
since
→
E
=Ex
ˆ
i
+Ey
ˆ
j
and we're given that
→
E
=(1.80
ˆ
i
+5.00
ˆ
j
)×105N/Cq=
(0.02kg)(9.8m/s2)
(1.80×105)cot37∘+5.00×105
q=2.65×10−7 C
B.)
T =
Fex
sinθ
=
qEx
sin37∘
T=0.079 N
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