1. Answer the following questions :
1. The distance between two fixed positive charges 4e and e isl. How should a
third charge q be arranged for it to be in equilibrium ? Under what conditions
will the equilibrium of the charge q be stabel and when will it be unstable ?
2. Two free positive charges 4e and e are a distance a apart. What change is
needed to acheive equilibrium for the entire system and where should it be
placed ?
3. A negative point charge 2e and a positive charge e are fixed at a distance l from
each other. Where should a positive test charge q be placed on the line
connecting the charges for it to be in equilibrium ? What is the nature of
equilibrium of the test charge with respect to longitudinal motions. Plot the
dependence of the force acting on this charge on the distance between it and the
charge te.
A charged particle is free to move in an electric field. It will travel
5. Two point charges Q and -3Q are placed at some distance apart. If the electric
field at the location of Q is E then at the locality of 3Q, it is
Answers
Answer:
Explanation:
3.Let the charge Q be placed at distance x from charge 2q. Now for the charge to be in equilibrium the forces due to both charges must cancel each other or the forces must be equal and opposite. Now the forces due to each charges is given by
F
2q
=2kQq/x
2
and
F
q
=kQq/(l−x)
2
for equilibrium we must have
F
2q
=F
q
or,
2kQq/x
2
=kQq/(l−x)
2
or,
(l−x)
2
x
2
=2 or,
l−x
x
=
2
or, x=
1+
2
2
l
l−x
x
=−
2
or, x=
2
−1
2
l
Since second solution of x lies outside the l we will neglect it. Hence the answer is-
$x=
2
+1
2
l
$
Now if the charge is displaced longitudinally towards charge q then x will increase hence F
2q
will decrease and F
q
will increase. Hence the charge will be forced to move towards charge q.
Similarly if it is moved towards charge 2q, F
2q
will increase and F
q
will decrease thus the charge Q will be moved towards charge 2q.
Hence the charge Q is in unstable equilibrium with respect to longitudinal motions.
2.Let"r" be the distance from the charge q where Q is in equilibrium .
Total Force acting on Charge q and 4q:
F=kqQ/r² + k4qQ/(l-r)²
For Q to be in equilibrium , F should be equated to zero.
kqQ/r² + k4qQ/(l-r)²=0
(l-r)²=4r²
⇒l-r=2r
⇒l=3r
⇒r=l/3
Taking the third charge to be -Q (say) and then on applying the condition of equilibrium on + q charge
kQ/(L/3)² =k(4q)/L²
9kQ/L²=4kq/L²
9Q=4q
Q=4q/9
Therefore a point charge -4q/9 should be placed at a distance of L/3 rightwards from the point charge +q on the line joining the 2 charges.
1.Given,
qA = +4e
qB = +e
AB = x
Let, Q be the third point charge located at some point P.
AP = x
therefore, PB = (a-x)
Force on Q due to qA,
F1 = kqAQ/x2
Force on Q due to qB
F2= kQqB/(a-x)2
Since the charge is in equilibrium,
F1 = F2
kqAQ/x2 = k qBQ/(a-x)2
4e/x2 = e/(a-x)2
4/x2 = 1/(a-x)2 (Taking square root on both sides)
2/x = 1/(a-x)
2(a-x) = x(1)
2a - 2x = x
2a = 3x
x = 2/3a
Therefore, the third point charge Q should be placed 2/3a from A.
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