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was all fun.”

– Thomas A. Edison

Questions in the
section ‘electrostatics’ were discussed on many occasions earlier on this site.
You may access them by clicking on the label ‘electrostatics’ below this post.
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Today we will discuss
a few more multiple choice practice questions on electrostatics:

(1) Point charges +

*q*and –3*q*are arranged at points A and B on the x-axis (Fig.). At which point (or points) is the electric field due to this system of charges zero?
(a) At two points on
the

*x*axis, one point to the right of the charge –3*q*and the other to the left of the charge +*q*.
(c) Somewhere on the

*x*axis to the right of the charge –3*q*.
(d) Somewhere on the

*x*axis, in between the two charges.
(e) The electric
field cannot be zero anywhere.

The electric field
lines due to a positive charge proceed radially outwards (diverge) from the
charge where as the electric field lines due to a negative charge proceed radially
inwards (converge) to the charge. At points on the x-axis lying in between A
and B the fields due to these charges are directed along the positive
x-direction. Therefore, the fields add up and the resultant field cannot be
zero. But at points on the x-axis to the left of the charge +

*q*, the fields are in opposite directions (along the negative x-direction due to the charge +*q*and along the positive x-direction due to the charge –3*q*) so that they can cancel at a point where the magnitudes of the fields due to the charges are the same. So the correct option is (b).
[At points on the
x-axis to the right of the charge –3

*q*, the fields are in opposite directions (along the positive x-direction due to the charge +*q*and along the negative x-direction due to the charge –3*q*. But the two fields cannot cancel anywhere here since the magnitude of the field due to the charge –3*q*is greater than that due to the charge +*q*(since the charge –3*q*is*nearer*as far as points on the right of B are concerned]
(2) A large flat
plate is positively charged so that it has uniform surface charge density σ. If
the electric field near the central region of the plate at a distance of 1 cm from
the plate is 9 NC

^{–1}, what will be the electric field at a distance of 3 cm from the plate?
(a) 81 NC

^{–1}
(b) 27 NC

^{–1}
(c) 9 NC

^{–1}
(d) 3 NC

^{–1}
(e) 1 NC

^{–1}
The electric field
due to the surface charge on the plate is normal to the surface of the plate.
Since the plate is large, the electric field near the central region of the
plate is uniform (with the electric field lines proceeding normal to the flat
surface and therefore parallel). This means that the electric field at a point
is

*independent of the distance*of the point from the plate so that electric field at 3 cm from the plate is 9 NC^{–1}itself.
[For AP Physics C
aspirants:

To find the electric
field near the surface of the charged plate, we can apply Gauss theorem and
accordingly imagine a Gaussian surface shaped as a rectangular parallelepiped
of cross section area A (Fig.). The electric field due to the surface charge on
the plate is normal to the surface of the plate. Therefore the electric field
is directed normal to the end faces of the rectangular parallelepiped so that
the total electric flux through the closed suface (rectangular parallelepiped)
is 2

*EA*where*E*is the magnitude of the electric field. (The flux through the side surfaces of the parallelepiped is zero since the electric field is parallel to these surfaces). The total charge enclosed by the closed surface is σ*A*. Therefore, by Gauss theorem we have
2

Therefore *EA =*σ*A/*ε_{0}where ε_{0}is the permittivity of free space.*E =*σ

*/*2

*ε*, which is

_{0}*independent*of distance]

(3) Charges +

*q*, –*q*and –*q*are placed at the vertices of an equilateral triangle ABC as indicated in the adjoining figure. What is the direction of the net electric field at the central point P?
(a) Along AP

(b) Along PA

(c) Along CP

(d) Along BP

(e) There is no field at P

The direction of
the electric fiel at P is the direction of the force acting on a test

*positive*charge placed at P. The charge +*q*will exert a*repulsive*force on the test positive charge. This is irected along AP. The charges –*q*and –*q*at B and C will exert attractive forces along PB and PC respectively. These two forces being of equal magnitudes, their resultant will be directed along AP. Therefore the resultant electric field due to all the three charges will be along AP [Option (a)].
(4) Points 1, 2 and 3
lie on the axis of an electric dipole AB. Point 4 lie on the equatorial line
(perpendicular bisector of AB) of the dipole. Out of the following choices
which one correctly gives the directions of the electric fields at the points
1, 2, 3 and 4?

When you place a

*positive*test charge at point (1) the net force on it is repulsive since the charge +*q*of the dipole is nearer than the charge –*q*. The direction of the force (and the electric field) is therefore*leftwards*.
At point (2) the
charge +

*q*exerts a repulsive force on the positive test charge where as the charge –*q*exerts an attractive force on it. Thus both forces act rightwards an hence the electric field at point (2) acts*rightwards*.
At point (3) the
net force on the positive test charge is attractive since the charge –

*q*of the dipole is nearer than the charge +*q*. The electric field at point (3) is therefore*leftwards*.
At point (4) the
positive test charge is repelled by the
charge +

*q*and attracted by the charge –*q*. these forces are equal in magnitude and are inclined equally upwards and downwards respectively. Their resultant acts rightwards and hence the electric field at point (4) is*rightwards*.
The above facts are
correctly given in option (e).

If you have a mental
picture of the distribution of the electric field lines due to an electric
dipole (See the figure below), you will be able to answer the above question in
no time].

(5) In the above
question, in the case of the points (1), (2), (3) and (4), where do you find
zero electric potential?

(a) Nowhere

(b) At point (1)

(c) At points (1), (2) and (3)

(d) At points (2) and (4)

(e) At point (2)

Points (2) and (4)
are equidistant from the charges +

*q*and –*q*. These equal and opposite charges produce positive and negative potentials of equal values so that they cancel each other, giving rise to zero potential at points (2) and (4).
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