Three point charges q, 2q and 8q are to be placed on a 9 cm 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 position of the charge q due to the other two charges?
Answers
Explanation:
Let the potential be 0 at infinity. Let's make an ansatz that the 2q and 3q charges are at the opposite end of the line; this would be the minimum potential energy positions if the q charge was not included, and switching the q charge with either of the larger charges would only increase the potential energy.
Let r be the distance of theq charge from the 2q charge. Then the potential energy of the system is:
U=k(2q2r+3q29−r+6q29)
We can minimize this by taking the derivitive with respect to r and setting it equal to 0:
0=kq2(−2r2+3(9−r)2)
r22=(9−r)23.
This has 2 solutions. However, we are only interested in solutions that lie along the 9cm line (solutions where r is positive), so r=9(6–√−2), which is indeed on the line and fulfills our intuition that the q charge should be closer to the 2q charge than the 3q charge.
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