Two point charges are separated by a distance of 10 cm. Charge on point A =+9 uC and charge
on point B = -4 uC. k = 9 x 109 Nm2C-2, 1 uC = 10-6 C. What is the change in electric potential
energy of charge on point B if accelerated to point A ?
Answers
Answered by
1
Answer:
3
Explanation:
Let
E
1
&
E
2
are the values of electric field due to q
1
and q
2
respectively magnitude of E
2
=
4π∈
0
1
r
2
q
2
E
2
=
(4
2
+3
2
)
9×10
9
×(25)×10
−6
V/m
E
2
=9×10
3
V/m
∴
E
2
=9×10
3
(cosθ
2
i
^
−sinθ
2
j
^
)
∴tanθ
2
=
4
3
∴
E
2
=9×20
3
(
5
4
i
^
−
5
3
j
^
)=(72
i
^
−54
j
^
)×10
2
Magnitude of E
1
=
4π∈
0
1
(1
2
+3
2
)
10
×10
−6
=(9×10
9
)×
10
×10
−7
=9
10
×10
2
∴
E
1
=9
10
×10
2
[cosθ
1
(−
i
^
)+sinθ
1
j
^
]
∴tanθ
1
=3
E
1
=9×
10
×10
2
[
10
1
(−
i
^
)+
10
3
j
^
]
E
1
=9×10
2
[−
i
^
+3
j
^
]=[−9
i
^
+27
j
^
]10
2
therefore
E
=
E
1
+
E
2
=(63
i
^
−27
j
^
)×10
2
V/m
∴ correct answer is (3)
Similar questions