two charges 3uc and 2uc are placed 1m a part in the air .calculate the colmbo force between them
Answers
Answer:
The situation is represented in the given figure. O is the mid-point of line AB.
Distance between the two charges, AB=20cm
∴AO=OB=10cm
Net electric field at point O=E
Electric field at point O caused by +3 C charge,
E
1
=
4π∈
0
(AO)
2
3×10
−6
=
4π∈
0
(10×10
−2
)
2
3×10
−6
N/C along OB
Where,
∈
0
= Permittivity of free space
4π∈
0
1
=9×10
9
Nm
2
C
−2
Magnitude of electric field at point O caused by - 3 C charge,
E
2
=
∣
∣
∣
∣
∣
4π∈
0
(OB)
2
−3×10
−6
∣
∣
∣
∣
∣
=
4π∈
0
(10×10
−2
)
2
3×10
−6
N/C along OB
∴E=E
1
+E
2
=2×[(9×10
9
)×
(10×10
−2
)
2
3×10
−6
] [As E
1
= E
2
, the value is multiplied with 2]
=5.4×10
6
N/ C along OB
Therefore, the electric field at mid-point O is 5.4×10
6
N/ C along OB.
Explanation:
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Answer:
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