Three identical charges q is kept on corners of equilateral triangle of side. Then calculate net force on any corner of triangle?
Answers
Answer:
Explanation:
Generally coulomb force between two charges q
1
and q
2
is
F
=
4πϵ
0
r
3
q
1
q
2
r
where r is the distance between the charges.
From figure, BC=
a
2
−
4
a
2
=
2
3
a
Let
F
O
be the coulomb force on charge at O due to rest charges at A(a,0) and B(
2
a
,
2
3
a)
thus,
F
O
=
4πϵ
0
a
3
q
2
(a
i
^
)+
4πϵ
0
a
3
q
2
(
2
a
i
^
+
2
3
a
j
^
)=
4πϵ
0
a
2
q
2
(
2
3
i
^
+
2
3
j
^
)
∴∣
F
O
∣=
(
4πϵ
0
a
2
q
2
)
2
(
4
9
+
4
3
)
=
4πϵ
0
1
.
a
2
3
q
2
Generally coulomb force between two charges q
1
and q
2
is
F
=
4πϵ
0
r
3
q
1
q
2
r
where r is the distance between the charges.
From figure, BC=
a
2
−
4
a
2
=
2
3
a
Let
F
O
be the coulomb force on charge at O due to rest charges at A(a,0) and B(
2
a
,
2
3
a)
thus,
F
O
=
4πϵ
0
a
3
q
2
(a
i
^
)+
4πϵ
0
a
3
q
2
(
2
a
i
^
+
2
3
a
j
^
)=
4πϵ
0
a
2
q
2
(
2
3
i
^
+
2
3
j
^
)
∴∣
F
O
∣=
(
4πϵ
0
a
2
q
2
)
2
(
4
9
+
4
3
)
=
4πϵ
0
1
.
a
2
3
q
2
Generally coulomb force between two charges q
1
and q
2
is
F
=
4πϵ
0
r
3
q
1
q
2
r
where r is the distance between the charges.
From figure, BC=
a
2
−
4
a
2
=
2
3
a
Let
F
O
be the coulomb force on charge at O due to rest charges at A(a,0) and B(
2
a
,
2
3
a)
thus,
F
O
=
4πϵ
0
a
3
q
2
(a
i
^
)+
4πϵ
0
a
3
q
2
(
2
a
i
^
+
2
3
a
j
^
)=
4πϵ
0
a
2
q
2
(
2
3
i
^
+
2
3
j
^
)
∴∣
F
O
∣=
(
4πϵ
0
a
2
q
2
)
2
(
4
9
+
4
3
)
=
4πϵ
0
1
.
a
2
3
q
2
Generally coulomb force between two charges q
1
and q
2
is
F
=
4πϵ
0
r
3
q
1
q
2
r
where r is the distance between the charges.
From figure, BC=
a
2
−
4
a
2
=
2
3
a
Let
F
O
be the coulomb force on charge at O due to rest charges at A(a,0) and B(
2
a
,
2
3
a)
thus,
F
O
=
4πϵ
0
a
3
q
2
(a
i
^
)+
4πϵ
0
a
3
q
2
(
2
a
i
^
+
2
3
a
j
^
)=
4πϵ
0
a
2
q
2
(
2
3
i
^
+
2
3
j
^
)
∴∣
F
O
∣=
(
4πϵ
0
a
2
q
2
)
2
(
4
9
+
4
3
)
=
4πϵ
0
1
.
a
2
3
q
2
Generally coulomb force between two charges q
1
and q
2
is
F
=
4πϵ
0
r
3
q
1
q
2
r
where r is the distance between the charges.
From figure, BC=
a
2
−
4
a
2
=
2
3
a
Let
F
O
be the coulomb force on charge at O due to rest charges at A(a,0) and B(
2
a
,
2
3
a)
thus,
F
O
=
4πϵ
0
a
3
q
2
(a
i
^
)+
4πϵ
0
a
3
q
2
(
2
a
i
^
+
2
3
a
j
^
)=
4πϵ
0
a
2
q
2
(
2
3
i
^
+
2
3
j
^
)
∴∣
F
O
∣=
(
4πϵ
0
a
2
q
2
)
2
(
4
9
+
4
3
)
=
4πϵ
0
1
.
2
3
q
2