“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 = √(AC

^{2}– BC^{2}) = √(0.5^{2}– 0.3^{2}) = 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?
(a)

*Q/ε*_{0}a
(b)

*Qa/ε*_{0}
(c)

*Q/*6*ε*_{0}
(d)

*Q/*8*ε*_{0}
(e)

*Qε*6_{0}/
The electric flux over a closed surface, according to
Gauss theorem, is

*Q*/ε_{0}where*Q*is the net charge*enclosed*by the surface and ε_{0}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/*4πε_{0}*r*^{2}. 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 4π*r*^{2}the electric field at any point on the surface must be*Ф/4πr*where^{2}*Ф*is the total electric flux produced by the charge*Q*. Therefore we have*Q/*4πε

_{0}

*r*

^{2}=

*Ф/4πr*

^{2}
This gives

*Ф = Q/ε*as stated in Gauss law.]_{0}
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/*6*ε*_{0}
(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 C

_{1}, C_{2}, C_{3 }and C_{4 }are of values satisfying the condition C_{1}/C_{2}= C_{3}/C_{4}, the junction of C_{1}and C_{2}is at the same potential as the junction of C_{3}and C_{4}(as in the case of a balanced Wheatstone brige). Therefore the capacitors C_{5}and C_{6}connected across the diagonal of the bridge have no effect and can be ignored.
The network
therefore simplifies to the series combination of C

_{1}and C_{2}connected in parallel with the series combination of C_{3}and C_{4}.
Series combined
value of C

_{1}and C_{2}= (3×6)/(3+6) = 2 μF
Series combined
value of C

_{3}and C_{4}= (1×2)/(1+2) = 2/3 μF
The capacitors C

_{1}and C_{4 }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 C

_{2}, C_{3}and C_{5}is connected to terminal A. The other plate of C_{2}as well as C_{3}is connected to terminal B while that of C_{5}is connected 6through C_{6}to terminal B. The series combination of C_{5}and C_{6}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.

## No comments:

## Post a Comment