A 1.45 x 10-3 kg mass of body is placed in the electric field of value 650 N/C. what is the sign and magnitude of electric charge for it to remain at rest in electric field. What value of electric field at which electric force becomes equal to weight of positive point charge.
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Answers
Answer:
some comtent is not related to the question but you will 3asily get the answer .
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Explanation:
Here, m=10−3kg,q=1.6×10−19C
Force on water drop due to electric field = weight of water drop
qE=mg
∴E=mgq=10−3×9.81.6×10−19=6⋅125×1016N/C
The electric field \vec E
E
E, with, vector, on top is a vector quantity that exists at every point in space. The electric field at a location indicates the force that would act on a unit positive test charge if placed at that location.
The electric field is related to the electric force that acts on an arbitrary charge qqq by,
\vec E = \dfrac{\vec F}{q}
E
=
q
F
E, with, vector, on top, equals, start fraction, F, with, vector, on top, divided by, q, end fraction
The dimensions of electric field are newtons/coulomb, \text{N/C}N/Cstart text, N, slash, C, end text.
We can express the electric force in terms of electric field,
\vec F = q\vec E
F
=q
E
F, with, vector, on top, equals, q, E, with, vector, on top
For a positive qqq, the electric field vector points in the same direction as the force vector.
COULOMB’S LAW
F
=
k
∣
q
1
q
2
∣
r
2
Coulomb’s law calculates the magnitude of the force F between two point charges, q1 and q2, separated by a distance r. In SI units, the constant k is equal to
k
=
8.988
×
10
9
N
⋅
m
2
C
2
≈
8.99
×
10
9
N
⋅
m
2
C
2
The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the two charges.