Three charges are at the corners of an equilateral triangle,
as shown in Fig. (1). (a) Calculate the electric field at the
position of the 2 μC charge due to the7 μC and 4μC charges. (b)
Use your answer to part (a) to determine the force on the 2 μC
charge
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Answer:
Answer
The force exerted on the 7.00−μC charge by the 2.00−μC charge is
F
1
=k
e
r
2
q
1
q
2
r
^
=
(0.500m)
2
(8.99×10
9
N.m
2
/C
2
)(7.00×10
−6
C)(2.00×10
−6
C)
×(cos60
o
i
^
+sin60
o
j
^
)
F
1
=(0.252
i
^
+0.436
j
^
)N
Similarly, the force on the 7.00μC charge by the −4.00−μC charge is
F
2
=k
e
r
2
q
1
q
3
r
^
=
(0.500m)
2
(8.99×10
9
N.m
2
/C
2
)(7.00×10
−6
C)(−4.00×10
−6
C)
×(cos60
o
i
^
−sin60
o
j
^
)
F
1
=(0.503
i
^
−0.872
j
^
)N
Thus, the total force on the 7.00−μC charge is
F
=
F
1
+
F
2
=(0.755
i
^
−0.436
j
^
)N
We can also write the total force as:
F
=(0.755N)
i
^
−(0.436N)
j
^
=
0.872Nat an angle of 330
o
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