find the position of neutral point along the line joining the two point charge Qa=-40mC and Qb=20mC which are 2cm apart in air
Answers
Question :-
Two point charges q
A
=3μC and q
B
=−3μC are located 20 cm apart in vacuum
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5×10
−9
C is placed at this point, what is the force experienced by the test charge?
Answer with explanation :-
(a) 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.
(b) A test charge of amount q=1.5×10
−9
C is placed at mid-point O.
q=1.5×10
−9
C
Force experienced by the test charge =F
∴F=qE
=1.5×10
−9
×5.4×10
6
=8.1×10
−3
N
The force is along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.
Hence, the force experienced by the test charge is 8.1×10
−3
Nalong OA.
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