Charge Q1=10nC is if the origin charge Q2=16 nC is on the X-axis at X=4m and charge Q3=20nC is at a point X=4m and Y=4m.find the magnitude and direction of the resultant force on Q2. kindly give me answer.
Answers
Answer:
Correct option is
C
(1.1×10
5
i
^
+2.5×10
5
j
^
) N/C
Diagram:
The total electric field E at P equals to the vector sum E
1
and E
2
, where E
1
is the field due to positive charge q
1
(7μC) at origin and E
2
is the field due to negative charge (−5μC)q
2
.
Magnitude of fields E
1
and E
2
are to be calculated as follows:
E
1
=
r
1
2
k∣q
1
∣
=9×10
9
×
(0.40)
2
7×10
−6
=
16
63×10
3
×10
2
=3.93×10
5
N/C
E
2
=
r
2
2
k∣q
2
∣
=9×10
9
×
(0.50)
2
5×10
−6
=1.8×10
5
N/C
[From using Pythagoras theorem, distance (r
2
) between q
2
and point P, r
2
=
(0.30)
2
+(0.40)
2
=0.50m]
Now, from the figure,
sinθ=
0.50
0.40
=
5
4
cosθ=
0.50
0.30
=
5
3
The vector E
1
has only y component.
The vector E
2
has x component given by E
2
cosθ=
5
3
E
2
and a negative y component given by −E
2
sinθ=
5
−4
E
2
.
Hence we can express vectors as:
E
1
=3.93×10
5
j
N/C
E
2
=[1.08
i
+(−1.44
j
)]×10
5
N/C
E
=
E
1
+
E
2
=[1.08×10
5
i
+(3.9−1.44)×10
5
j
]N/C
=[1.1×10
5
i
+2.49×10
5
j
]N/C
=[1.1×10
5
i
+2.5×10
5
j
]N/