“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

*C*_{1}and*C*_{2}(let us say) having values 6 μF and 2 μF. If the maximum voltage that the parallel combination can withstand is*V*_{max}, the voltage*V*_{1}across the 6 μF capacitor on applying this voltage across the series combination is given by*V*

_{1}=

*V*

_{max}

*C*

_{2}/(

*C*

_{1}+

*C*

_{2})

[The charge

*Q*on each capacitor on connecting the voltage*V*_{max}across the series combination is given by*Q*=*C*_{1}*C*_{2}*V*_{max}/(*C*_{1}+*C*_{2}), remembering that the effective capacitance of the series combination is*C*_{1}*C*_{2}/(*C*_{1}+*C*_{2})].
Therefore,

*V*_{1}=*V*_{max}×2/(6+2) =*V*_{max}/4
Since

*C*_{1}can withstand a maximum voltage of 3 kV we have*V*_{max}/4 = 3 kV
This gives

*V*_{max}= 12 kV.
[Do not jump to a conclusion at this stage, You
have to check whether

*C*_{2}will be intact on applying the above 12 kV across the series combination].
The voltage

*V*_{2}across the 2 μF capacitor on applying this voltage*V*_{max}across the series combination is given by*V*

_{2}=

*V*

_{max}

*C*

_{1}/(

*C*

_{1}+

*C*

_{2})

Threfore

*V*_{2}=*V*_{max}×6/(6+2) = 6*V*_{max}/8
Since

*C*_{2}can withstand a maximum voltage of 6 kV we have 6*V*_{max}/8 = 6 kV
This gives

*V*_{max}= 8 kV.
This value of

*V*_{max}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*
(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 =*(1/4πε

_{0}) (

*q*

^{2}/

*d*

^{2}) where ε

_{0}is the permittivity of free space (and air, very nearly),

*q*is the charge on each sphere and

*d*is the distance between the spheres.

When the spheres are in the oil the
electrostatic force

*F*_{1}*between the spheres is given by**F*

_{1}= (1/4πε

_{0}

*K*) (

*q*

^{2}/

*d*

^{2}) where

*K*is the dielectric constant of the oil.

The
real weight

*W*of each sphere is given by*W*=

*V*

*ρg*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

*W*_{1}of*each sphere when immersed in oil is given by**W*

_{1}=

*V*

*ρg*–

*V*

*σ*

*g*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 =*(1/4πε_{0}) (*q*^{2}/*d*^{2})/*V**ρg*………………..(i)
When the spheres are in oil, we have

tan α =

*F*_{1}*/W*_{1}*=*(1/4πε_{0}*K*) (*q*^{2}/*d*^{2}) / (*V**ρg*–*V**σ**g*)……(ii)
Dividing
Eq. (i) by Eq. (ii) we have

1 =

*K*(*ρ*–*σ*) /*ρ*
Therefore,

*K =**ρ/*(*ρ*–*σ*) = 1200/400 = 3