“Example isn't
another way to teach, it is the only way to teach.”
– Albert Einstein
Today we will discuss a few questions from
electrostatics. You will find many questions (with solution) in this section
discussed earlier on this site. You can access all those questions by clicking
on the label ‘electrostatics’ below this post.
(1) A 6 μF capacitor is connected in series with a 2 μF
capacitor. The 6 μF capacitor can
withstand a maximum voltage of 3 kV where as the 2 μF capacitor can withstand a
maximum voltage of 6 kV. The maximum voltage that the parallel combination can
withstand is
(a) 2 kV
(b) 3 kV
(c) 6 kV
(d) 8 kV
(e) 12 kV
We have capacitors C1 and C2
(let us say) having values 6 μF and 2 μF. If the maximum voltage
that the parallel combination can withstand is Vmax, the voltage V1
across the 6 μF capacitor on applying this voltage across the series combination is
given by
V1 = Vmax C2/(C1 + C2)
[The charge Q
on each capacitor on connecting the voltage Vmax
across the series combination is given by Q
= C1C2 Vmax/(C1 + C2), remembering that the effective capacitance of the
series combination is C1C2/(C1 + C2)].
Therefore, V1
= Vmax×2/(6+2) = Vmax/4
Since C1
can withstand a maximum voltage of 3 kV we have Vmax/4 = 3 kV
This gives Vmax = 12 kV.
[Do not jump to a conclusion at this stage, You
have to check whether C2
will be intact on applying the above 12 kV across the series combination].
The voltage V2
across the 2 μF capacitor on applying this voltage Vmax across the series combination is given by
V2 = Vmax C1/(C1 + C2)
Threfore V2 = Vmax×6/(6+2) = 6 Vmax/8
Since C2
can withstand a maximum voltage of 6 kV we have 6 Vmax/8 = 6 kV
This gives Vmax = 8 kV.
This value of Vmax being lower than that
obtained (12 kV) on considering the 6 μF capacitor, the correct option is 8 kV.
(2) A uniform electric field of intensity E newton/coulomb directed along the
positive x-direction exists in a region of space (Fig.). The x- direction is
horizontal. A, B, C and D are points at the corners of a square of side a, with AB and CD parallel to the x-direction. If the electric
potential at point A is V volt, what
is the potential (in volt) at the point D?
(a) V
(b) V – aE
(c) V – √2 aE
(d) V + √2 aE
(e) V + aE
Since the electric
field acts along the positive x-direction, the potential decreases as we move along the positive x-direction.
[Remember that the
electric field is the negative gradient
of potential].
While moving from A
to D the x-coordinate increases by ‘a’ and hence the potential decreases by aE. Therefore, the potential at D is V – aE [Option (b)].
(3) In the above
question what is the potential difference between points A and C?
(a) V
(b) V – aE
(c) V + aE
(d) aE
(e) Zero
Since the electric
field acts along the x-direction, the potential will change only if the
x-coordinate changes. Points A and C have the same x-coordinates and hence they
are at the same potential. Therefore, the potential difference between points A
and C is zero.
(4) Two small identical spheres are charged equally and suspended in air by
strings of equal length. The strings make a small angle θ with each other (Fig.). When the spheres are
immersed in oil of density 800 kg m–3 the angle between the strings is found to be unaltered. If
the density of the material of the spheres is 1200 kg m–3, what is the dielectric constant of the oil?
(a) 1.5
(b) 2.5
(c) 3
(d) 3.5
(e) 4
The repulsive electrostatic force F between the spheres in air is given by
F = (1/4πε0) (q2/d2) where ε0 is
the permittivity of free space (and air, very nearly), qis the charge on each sphere and d is the distance between the spheres.
When the spheres are in the oil the
electrostatic force F1 between the spheres is given by
F1
= (1/4πε0K) (q2/d2) where K is the dielectric constant of the oil.
The
real weight W of each sphere is given
by
W
= Vρg where V is the volume, ρ is the density of the material of the sphere and
g
is the acceleration due to gravity.
The
apparent weight W1 of each sphere when immersed in oil is
given by
W1
= Vρg
– Vσg where σ is the density of the oil.
[Note
that Vσg is the upthrust or the force of
buoyancy due to the oil]
When
the spheres are in air, we have (Fig.)
tan α = F /W
= (1/4πε0) (q2/d2)/ Vρg………………..(i)
When the spheres are in oil, we have
tan α = F1 /W1 = (1/4πε0K) (q2/d2) / (Vρg – Vσg)……(ii)
Dividing
Eq. (i) by Eq. (ii) we have
1 = K(ρ – σ) / ρ
Therefore, K = ρ/(ρ – σ) = 1200/400 = 3
