Three point charges q, 2q and 8q are to be placed on a 18cm long straight line.
Find the positions where the charges should be placed such that the potential
energy of this system is minimum. In this situation, what is the electric field at
the charge q due to the other two charges?
Answers
Answer:
ANSWER
Given q, 2q, 8q three charges present,now the system potential energy to be minimum we will place q charge in between 2q & 8q.
Such that,
Explanation:
ANSWER
Given q, 2q, 8q three charges present,now the system potential energy to be minimum we will place q charge in between 2q & 8q.
Such that,
Now, The potential energy of the system is given by,
u=
r
12
Kq
1
q
2
+
r
23
Kq
2
q
3
+
r
31
Kq
3
q
1
u=K[
x
2q
2
+
9−x
8q
2
+
9
16q
2
]
⇒u=Kq
2
[
x
2
+
9−x
8
+
9
16
]
Now potential energy of the system depends on the value of x.
∴u
1
=
x
2
+
9−x
8
{u : minimumizing differentiating w.r.t x}
will give
dx
du
1
=
x
2
−2
+
(9−x)
2
8
for u
1
to be minimum,
dx
du
1
=0⇒
x
2
−2
+
(9−x)
2
8
=0
x
2
1
=
(9−x)
2
4
⇒ x=3cm or −9cm
Now, x=−9cm is not applicable hence we will take x=3cm
Potential energy at x=3,
u/
x=3
=Kq
2
[
3
2
+
6
8
+
9
16
]=
3
Kq
2
[6+
3
16
]=
9
34Kq
2
U
The minimum value of P.E=
9
34Kq
2
In this situation electric field at charge q is given by,
E
=
E
1
+
E
2
where,
E
1
=
(3)
2
K2q
&
E
2
=
(6)
2
K8q
=
9
2Kq
(
i
^
) =
9
2Kq
(−i)
as
E
1
&
E
2
are opposite directed resultant is given by,
E
=
9
2Kq
−
9
2Kq
=0