“Iron rusts from disuse, stagnant water
loses its purity and in cold weather becomes frozen; so does inaction sap the
vigors of the mind.”
– Leonardo da Vinci
Questions on electrostatics
were discussed on many occasions on this site. You can access them by clicking on
the label ‘electrostatics’ below this post or by trying a search for
‘electrostatics’ using the search box provided on this page. Today we shall
discuss a few more questions in this section.
(1) Right angled
triangle ABC is located in a uniform electric field E. Sides AC and BC have
lengths 0.5 m and 0.3 m respectively and BC is at right angles to the electric
field lines (Fig.). If the electric potential difference between A and C is 80
V, what is the magnitude of the electric field E?
(a) 80 V/m
(b) 160 V/m
(c) 200 V/m
(d) 100 V/m
(e) 16 V/m
The electric potential at the point B is the same
as that at the point C (since the straight line BC is at rjght angles to the
direction of the uniform electric field E). Therefore the potential difference
between points A and B is 80 V. The electric field E is
directed along AB and hence the magnitude of E is (80 V)/(0.4 m) = 200 V/m.
[AB = √(AC2 – BC2) = √(0.52 – 0.32) = 0.4 m]
(2) In the above
question, what is the component of electric field along the direction AC?
(a) 80 V/m
(b) 160 V/m
(c) 200 V/m
(d) 100 V/m
(e) 40 V/m
The potential
difference between the points A and C is 80 V and the distance between these
points is 0.5 m. Therefore, the component of electric field along the direction
AC is (80 V)/(0.5 m) = 160 V/m.
(3) A cube of side a has a charge Q at its centre (Fig.). What is the electric flux through one 
face
of the cube?
face
of the cube?
(a) Q/ε0a
(b) Qa/ε0
(c) Q/6ε0
(d) Q/8ε0
(e) Qε0/6
The electric flux over a closed surface, according to
Gauss theorem, is Q/ε0 where
Q is the net charge enclosed by the surface and ε0
is the permittivity of free space.
[You can understand
the above even without knowing Gauss law. You know (from inverse square law)
that the electric field at distance r
from a point charge Q is Q/4πε0r2. The electric field at any point is the electric flux
through unit area held with the plane of the area perpendicular to the electric
field lines. Imagine a spherical surface of radius r such that the charge Q
is at the centre. Since the area of the spherical surface is 4πr2 the electric field at any
point on the surface must be Ф/4πr2
where Ф is the total electric flux
produced by the charge Q. Therefore we
have
Q/4πε0r2 = Ф/4πr2
This gives Ф = Q/ε0 as stated in Gauss
law.]
Since the charge Q is at the centre of the cube, all the
six faces of the cube receive equal electric flux so that the flux through one
face of the cube is Ф/6 = Q/6ε0
(4) The effective
capacitance between terminals A and B in the network shown in the adjoining
figure is
(a) 16 μF
(b) 8 μF
(c) 6 μF
(d) 16/3 μF
(e) 8/3 μF
Since the
capacitors C1, C2, C3 and C4 are of
values satisfying the condition C1/C2 = C3/C4,
the junction of C1 and C2 is at the same potential as the
junction of C3 and C4 (as in the case of a balanced
Wheatstone brige). Therefore the capacitors C5 and C6
connected across the diagonal of the bridge have no effect and can be ignored.
The network
therefore simplifies to the series combination of C1 and C2
connected in parallel with the series combination of C3 and C4.
Series combined
value of C1 and C2 = (3×6)/(3+6) = 2 μF
Series combined
value of C3 and C4 = (1×2)/(1+2) = 2/3 μF
Therefore the
effective capacitance between terminals A and B = 2 μF + (2/3) μF = 8/3 μF
The capacitors C1
and C4 in the above question are short circuited and the circuit then
gets modified as shown in the figure. What is the effective capacitance between
the terminals A and B in this situation?
(a) 11 μF
(b) 9 μF
(c) 8 μF
(d) 7.5 μF
(e) 6.5 μF
Have a careful look
at the circuit. You will find that one plate of C2, C3
and C5 is connected to terminal A. The other plate of C2
as well as C3 is connected to terminal B while that of C5
is connected 6through C6 to terminal B. The series combination of C5
and C6 gives an effective capacitance of 1μF. Thus we have three
capacitances 1μF, 1μF and 6μF connected in parallel across the terminals A and
B.
Therefore the
effective capacitance between terminals A and B in this case is 1μF + 1μF + 6μF
= 8μF.
