MPS/PHY-XII-2011/S1
PHYSICS
ASSIGNMENT-1 SOLUTIONS
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1.
Two equal charges, distance x apart, exert a force on one another. The charge on one of
the charges is doubled. What is the ratio of the distance between the two charges now
and earlier if the force in the two cases is same?
F=k
A.
2.
q ×q
q ×2q r'
2
= k ,2 ∴ =
2
r
r
r
1
A force F is acting between two charges placed some distance apart in vacuum. If a brass
rod is placed between these two charges, how does the force change?
A.
For any metal κ = ∞, ∴ Fbrass =
Fvac F
= = 0 . i.e., in the presence of brass rod, the force between
∞
κ
two charges becomes zero.
3.
Write the dimensional formula for the permittivity constant ε0 of free space.
A.
[M-1 L-3 T4 A2]
4.
If a charge +Q is fixed at a distance‘d’ in front of
an infinite metal plate, draw the
electric lines of force indicating the direction clearly.
A.
5.
Two point charges of charge values Q and q are placed at distances x and x/2
respectively from a third charge of charge value 4q, all charges being in the same
straight line. Calculate the magnitude and nature of charge Q such that the net force
experienced by the charge q is zero.
A.
Suppose the three charges are placed as shown in the figure
For the equilibrium of charge q, the charge Q must have the same sign as that of q or 4q; so
that the forces FA and FB are equal and opposite. As FA = FB
∴
6.
1 4q ×q
1 q ×Q
=
⇒ Q = 4q
4πε 0 (x /2)2 4πε 0 (x /2)2
Two identical charges, Q each are kept at a distance r from each other. A third charge q
is placed on the line joining the above two charges such that all the three charges are in
equilibrium. What is the magnitude, sign and position of the charge q?
A.
Suppose the three charges be placed in the manner as shown in the figure. The charge q will be
in equilibrium if the forces exerted on it by the charges at A and C are equal and opposite.
k
Qq
Qq
or x 2 = (r - x)2
=k
2
x
(r - x)2
or x = r-x or x = r/2 . Since the charge at A is repelled by the similar charge at C, so it will be
in equilibrium if it is attracted by the charge q at B, i.e., the sign of charge q should be
opposite to that of charge Q.
Force of repulsion b/w charges at A and C = Force of attraction b/w charges at A and B
k
7.
Qq
QQ
=k 2
2
(r /2)
r
or q = Q/4
An infinite number of charges each equal to 4 µC are placed along x-axis at x =1m,
x = 2m, x = 4m, x = 8m and so on. Find the total force on a charge of 1C placed at the
origin.
A.
Here q = 4µC = 4 X 10-6C, q0 = 1 C. By the principle of superposition, the total force acting on a
charge of 1 C placed at the origin is
F=
qq 0
4πε 0
1 1 1

1 1 1

9
-6
 2 + 2 + 2 +.... = 9×10 × 4×10 ×1  2 + 2 + 2 +... .
r
r
r
1
2
4


2
3
 1

Sum of the infinite geometric progression = a/(1-r) = 1 /(1-1/4) = 4/3
F = 9 X 109 X 4 X 10-6 X 4/3 = 4.8 X 104 N
8.
Two similar balls each having mass m and charge q are hung from a silk thread of length
1/3
 q 2l 
l, prove that equilibrium separation x = 
 when each thread makes a small
 2πε 0 mg 
angle θ with the vertical.
A.
According to coulomb’s law, the electrostatic repulsion between the two balls will be F = kq2/x2
For small angular displacement θ, we have sinθ≈ θ(rad) = arc/radius = (x/2)/l = x/2l
Restoring force F on each ball = mgsinθ ≈ mgx/2l
For equilibrium, mg
x
1 q2
or
=
2l 4πε 0 x 2
1/3
 q 2l 
q 2l
or x = 
x =

2πε 0 mg
 2πε 0 mg 
3
9.
Five point charges each of value +q coulomb are placed on five vertices of a regular
hexagon of side L meters. Find the magnitude of force on a charge -q coulomb placed at
the center of the hexagon.
The situation is shown in the following figure.
Forces on charge –q at O due to charges at A and D are equal
and opposite and hence cancel out.
Similarly, the forces due to charges at B and E cancel out.
The only force on charge –q at O is exerted by charge +q at C.
It is given by F = kqq/(OC)2 = kq2/L2 N along OC.
10.
Calculate the voltage needed to balance an oil drop carrying 10 electrons when located
between the plates of a capacitor which are 5 mm apart. The mass of oil drop is 3 X 10-16
kg. Take g = 10m/s2
Here q = 10e = 10 X 1.6 X 10-19C, d= 5mm = 5 X 10-3m, m= 3 X 10-16kg, g=10m/s2
When the drop is held stationary, Upward force on oil drop due to electric field = wt. of oil drop
qE = mg or q.(V/d) = mg. ∴ V =
11.
mgd 3×10-16 ×10×5×10-3
=
= 9.375V
q
10×1.6×10-19
An infinite number of charges, each equal to q are placed along x-axis at x =1, x =2,
x = 4,x = 8,… and so on. (i) Find the electric field at the point x = 0 due to this set up of
charges. (ii) What will be the electric field, if in the above set up, the consecutive charges
have opposite signs.
(i)At the point x = 0, the electric fields due to all the charges are in the same x-direction and
hence get added up E =
= E=
1
4πε 0
q q
q q

12 + 22 + 42 + 82 + ...


q  1 1
1
q
1
q

1+ + +
+ ... =
=

1
4πε 0  4 16 64
4πε 0 1 −
3πε 0

4
(ii) Electric field at x = 0 is E =
=E=
1 q q
q q

− 2 + 2 − 2 + ...
2

4πε 0 1 2
4 8

q
1
q  1 1
1
q

=
1− + −
+ ... =

1
4πε 0
4πε 0  4 16 64

 5πε 0

1 − 1 − 
4


12.
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the
electric intensity E at a point on the axis of the ring. Hence show that for points at large
distances from the ring, it behaves like a point charge.
Suppose that the ring is places with its plane perpendicular to the x-axis, as shown in the
above figure. Consider a small element dl of the ring.
As the total charge q is uniformly distributed the charge dq on the element dl is
dq =
r
∴ The magnitude of the field dE produced by the element dl at the field point P is
r
dq
kq dl
dE = k 2 =
r
2π a r 2
r
As shown in the figure, the field dE has two components:
1. the axial point dEcosθ
2. the perpendicular component dEsinθ
q
2π a
dl
Since the perpendicular components of any two diametrically opposite elements are equal and
opposite, they all cancel out in pairs. Only the axial components will add up to produce the
r
resultant field E at point P, which is given by,
2π a
E=
∫ dE cos θ
[as only the axial components contribute towards E ]
0
2π a
=
∫
0
=
or
kq dl x kqx 1
=
2π a r 2 r 2π a r 3
2π a
∫ dl
[ as cosθ = x/r ]
0
kqx 1 2π a kqx
1
[l ]0 =
2π a [as r2 = x2+a2 ]
3
2
2π a r
2π a ( x + a 2 )3/2
E=
kqx
1
qx
=
2 3/2
2
(x + a )
4πε 0 ( x + a 2 )3/2
2
Special case: For points at large distances from the ring x>>a
∴E =
kq
1 q
=
2
x
4πε 0 x 2
This is the same as the field due to a point charge, indicating that for far off axial points, the
charged ring behaves as a point charge.