Lecture 8
Multiple Choice Questions :
1.
A point charge -3Q lies at the centre of a conducting shell of radius 2R. The net charge
on the outer surface of the shell is
a. -3Q
b. Zero
c. +1.5 Q
d. +3Q
2. Two identical spherical conductors A and B of radius R , each carrying a charge Q are
kept at some distance from each other. A third spherical conductor C , initially uncharged, is
first brought into contact with A and then with B before finally being removed to a far away
distance. If the charge on C is (10/9)Q, the radius of C is
a. R/5
b. R/3
c. R/2
d. 2R
3. 64 identical spherical drops of mercury are combined to form a large drop. If the
potential of each smaller drop is 1 V, the potential of the final drop is
a. 1 V
b. 4 V
c. 16 V
d. 64 V
4. Infinite number of charges of equal magnitude Q are placed along the axis at distances
a, 2a, 3a, ….from the origin. If the charges alternate in sign and the charge closest to the
origin is positive, the potential at the origin due to the charge distribution is
a.
b.
c.
d. Zero
5. Three charges, Q, q and q are arranged at the vertices of a right angled isosceles triangle
of base a. If the charge Q are fixed, the configuration has minimum electrostatic energy
when q is equal to
Q
a. −𝑄𝑄2+12
b.
q
q
c.
d.
6. The electric potential in a region along the x axis varies with distance x (in meters) as
(Volts). The force acting on a 1µC charge located at
is
a.
N along the positive x axis
b.
N along the negative x axis
c.
N along the positive x axis
d.
N along the negative x axis
7. Four electric charges +q, +q, –q and –q are placed at the corners of a square of side 2L
(see figure). The electric potential at point A, midway between the two charges +q and +q,
is
a.
b.
c.
d. Zero
8. Two uniformly charged concentric rings of radii R and 2R are placed on a plane. Each
ring has a charge density λ. The electric potential at the centre of the rings is
a.
b.
c.
d.
9. A hemisphere of radius R is charged with a uniform surface charged density s on its
curved surface. The potential the centre is
a.
b.
c.
d.
10. A hollow metal sphere of radius R is charged to a potential of 10 V on its surface. What
is the potential at the centre of the sphere?
a.
b.
c.
d.
Zero
+10 V
-10 V
Same as its value outside at a distance of R from the surface of the sphere.
Answers to Multiple choice questions:
1. (a) 2. (d) 3. (c) 4. (c) 5. (c) 6. (d) 7. (b) 8. (c) (9) (b) 10. (b)
Problems
1. A uniformly charged sphere of radius R contains a charge Q. Choose the origin of
coordinate system at the centre of the sphere and let V(0)=0. Find an expression for the
potential both inside and outside the sphere.
2. Two identical metal plates have area 1 m2 each and are separated by 3 cm. Initially both
are uncharged. A charge of 2 nC is transferred from the plate on the left to the plate on the
right and equilibrium is allowed to be established. Neglecting edge effects, calculate the
electric field (a) in the region II between the plates at a distance of 0.5 cm from the plate to
the right and (b) at points to the immediate left of the plate on the left (region I) and at
points to the immediate right of the plate to the right in region III.
I
II
III
3. Two identical thin rings, each of radius R, are placed coaxially at a distance R from each
other. The rings carry charges Q 1 and Q 2 uniformly spread over the rings. Find the work
done in removing a charge q from the centre of the first ring to that of the second ring.
4. A metal sphere of radius R carrying a charge q is surrounded by a thick concentric
conducting shell of radii a and b. The shell has no charge. Calculate the charge densities on
the metal sphere as well as on the inside and outside surface of the shell. With the
reference of the potential at infinite distances, determine the potential at thecentre of the
sphere. If now, the outer surface of the shell is grounded, what will be the potential at the
centre of the sphere?
5. A metal sphere of radius R has two spherical cavities of radii a andb. The former has a
charge q at the centre of the cavity while the latter has a charge Q at its centre. Find the
charge densities in the interior of the two cavities and on the outside surface of the sphere.
Also determine the field outside the sphere at a distance r from the centre of the sphere.
6. Potential on the surface of a sphere is 400 V and its value drops to 100 V at a distance of
60 cm from the surface. Calculate the radius of the sphere and the amount of charge
contained in the sphere.
7. Two equal charges q are located at a distance 2d apart. Find the expression for the
electric potential at a distance of z along the perpendicular bisector of the line joining the
two charges. Using this determine the electric field at that point.
8. A metal sphere of radius R carries a charge Q. Surrounding this is a concentric metallic
shell of inner radius 2R and outer radius 3R. If the shell carries a total charge 3Q, find the
potential for r<R.
9. Can an electric field be given by the expression
?
10. How much work is done in moving a 0.5nC charge on the surface of a sphere of radius
to a final position
The electrostatic
2m from an initial position
field in the region is given by
(in V/m).
Hints for solutions to Problems
1. Electric field is known from Gauss’s law to be
for r <R and
for r>R. For r<R
,
. For r>R, take the corresponding
expression for the field for outside the sphere and integrate from the surface of the sphere
to
the
point
where
the
potential
is
to
be
found.
We
get
2. The charge on the left plate is -2nC and on the right plate is +2nC. Without assuming that
the final charges are on the faces of plate facing each other, we can assume that the charge
density on the left plate (plate 1) is σ 1L on its left face and σ 1R on its right face. Likewise, the
charge density on the right plate (plate 2) are σ 2L and σ 2R respectively. We also have
C/m2. Further,
.
For a point in region II,
σ1L σ1R
σ2L
σ2R
Substituting values, the field strength is approximately 226 N/C. By similar arguments, the
field to the left of the left plate and that to the right of the right plate can be shown to be
zero. (Solution becomes simpler if one assumes a- priori that the charges appear only on
the faces of the two plates that face each other.)
3. Calculate the potential at P due to the first ring which is
is at a distance
ring is
Similarly, since P
from all points on the second ring, the potential at P due to the second
. Thus the net potential at P is
potential at P’ is
. Similarly, the
. The work done is q times the potential difference
between P and P’, which is
.
P
P’
Q1
Q2
4. The charge densities are
. Inside the metal sphere (for r<R) the
field is zero. Field for R<r<a as well as for r>b, the field is inverse square and is given by
. The
potential
field
can
be
calculated
as
a
line
integral
of
the
electric
If the outside of the shell is grounded, the charge density on that surface becomes zero
while the charge densities on all other surfaces remain unchanged. The only non-zero field
is in the region R<r<a. Further V(b)=0. Thus
.
5. The charge densities on the inside surface of the cavities are
while the charge density on the outside surface is
,
. The field outside the sphere is same as
due to a charge q+Q located at the centre of the sphere.
6. R=0.2 m and
C.
7. Take the line joining the charges along the x-axis and the perpendicular bisector along the z axis.
Since the point at which the potential is to calculated is at a distance
charges, the potential is given by
from each of the
. The electric field is the gradient of the
potential and is given by
.
8. The potential in the region r<R is the same as that on the surface of the metal sphere. The charge
distribution on the shell will be –Q on the inner surface and 4Q on the outer surface. We can use
superposition principle to find the potential at r=R. This gives
. You
can also find it by determining the field distribution using Gauss’s law and determining potential
9. No, the curl the given field is not zero (calculate the curl explicitly).
10. Show that the field is conservative and obtain the corresponding potential. The potential is given
(in V). Then convert the given positions to Cartesian. The work done is equal to
by
q times the change in potential. Ans.