A charge Q is divided into two parts q1 and q2 such thatthey experience maximum force of repulsion when separated by a certain distance.the ratio of Q,q1 and q2 is
Answers
Answer:
Let initial separation between the particles be l and this remains constant during the motion. If relative separation remains constant then relative velocity should be zero; hence relative acceleration of the particles should also be zero.
a
r
=a
1
−a
2
=0
or
m
1
q
1
E−F
e
=
m
2
F
e
−q
2
E
or
m
1
m
2
(q
1
m
2
+q
2
m
1
)E
=F
e
[
m
1
1
+
m
2
1
]
or (q
1
m
2
+q
2
m
1
)E=F
e
(m
1
+m
2
)=
4πε
0
1
l
2
q
1
q
2
(m
1
+m
2
)
After solving we get, l=
2
1
πε
0
E(q
2
m
1
+q
1
m
2
)
q
1
q
2
(m
1
+m
2
)
.......jridoor
Answer:
Coulomb's law tells us that the force of repulsion is proportional to the product of the two charges and inversely proportional to the square of the distance between them.
F=kqF=kq_1q1q_2/r22/r2
If you substitute q1=Qq1=Q and q2=Q−qq2=Q−q for each case:
(A) F=(3k/4)Q2/r2F=(3k/4)Q2/r2
(B) F=(k/2)Q2/r2F=(k/2)Q2/r2
(C) F=0F=0
(D) F=(7k/8)Q2/r2F=(7k/8)Q2/r2
. Note that the force is positive for both positive and negative Q since the charges are multiplied.
You could find the maximum by taking the
first and second derivatives, but from simple inspection you can see that the force is simply 0 everywhere for (C), and for (A), (B), (D) force increases as distance approaches 0. This demonstrates that force increases with decreasing distance regardless of the magnitude of the two charges.