a charge+2microcoulomb is placed at a point A(2,1,0). another charge-3microcoulomb is placed at a point B(4,2,1). calculate net for ce on a charge +5microcoulomb placed at a point C(_1,3,2)
Answers
Answer:
Given :
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]
Explanation: