two point charges + 2 microcoulomb and + 6 microcoulomb are located at points A and B respectively then CM apart in vacuum 1.what is the electric field at a point O of the line AB
2.if a positive charge 5 nC is placed at a point O .what is the force experienced by that charge
Answers
Answer:
Q
1
=400μC=400×10
−6
C
Q
2
=100μC=100×10
−6
C
Distance between two charges d=60cm
Thus distance of point O from each charge r=
2
60
cm=30cm=30×10
−2
m
Consider that point charges are kept at point A and point B. If 'O' is the midpoint, where we need to calculate the electric field due to these charges.
At point O, electric field due to point charge kept at A,
E
1
=
4πϵ
0
1
×
r
2
Q
1
=9×10
9
×
(30×10
−2
)
2
400×10
−6
[in the direction of AO]
At point O, electric field due to point charge kept at B,
E
2
=
4πϵ
0
1
r
2
Q
2
=9×10
9
×
(30×10
−2
)
2
100×10
−6
[in the direction of BO]
So, resultant electric field
E
=
E
1
−
E
2
E
=
(30×10
−2
)
2
9×10
9
×10
−6
(400−100)
=
900×10
−4
9×10
9
×10
−6
×300
=3×10
7
N/C [in the direction of AO]
solution
Explanation: