OBC, etc. are equilateral triangles. Therefore,OA = OB = OC= OD = OE = OF
= r = 10cm = 10-1m,
As potential is scalar, therefore, potential at O is
The direction of electric field every point on this surface is along normal to the plane in the
(a) Inside the sphere, E = 0
(b) Just outside the sphere. (say, on the surface of the sphere )
For each capacitor, V is same = 100 volt
Capacity increases to C’ = KC0 = 6×1.77×10-11 F 6
Charge increases to q’ = C’ V = 6×1.77×10-11 ×102 C.
(b) After the supply was disconnected, charge remains constant. New capacity,
C’ KC0 = 6×1.77×10-11 F
q0 = -2 from P to Q,
where OP = rp = 3cm = 3
and OQ = rQ = 4 cm = 4 10-2 m
As electrostatic forces are conservative forces, work done is independent of the path. Therefore,
there is no relevance of point R.
Further, electric field intensity at the centre due to all the eight charges is zero, because the fields
due to individual charges cancel in pairs.
(b) Let P be the point in a plane normal to the line passing through the mid point, where OP =
along AP produced
Electrc intensity at P due to q2 ;
(b) Is the electric filed intensity inside a cavity (with no charge ) zero, even if the shell is not spherical, but has any irregular shape ? Explain.
shell induces charge – q on the inner surface and charge+ q on the outer surface of the shell, therefore. Total charge on inner surface of the shell is – q and total charge on the outer
surface of the shell is (Q + q).
(b) Electric field intensity inside a cavity with no charge is zero. Even when the shell has any irregular
shape. If we were to take a closed loop. Part of which is inside the cavity along a field line, and the
rest outside it. Then net work done by the field in carrying a test charge over the closed loop will
not ne zero. This is impossible for an electrostatic field. Hence electric field intensity inside a cavity
with no charge is always zero.
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
sheet of charge, on left side (side 1)
(b) To show that the tangential component of electrostatic field continuous from one side of a
charged surface to another, we use the fact that work done by electrostatic field on a closed closed
loop is zero.
A. The charge q = λ l spreads uniformly on the outer surface of A . It induces – q charge on the
cylinder B, which spreads on the inner surface of B. An electric field E ⃗ is produced in the space
between the two cylinders, which is directed radially outwards. Let us consider a co – axial
cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is
(a) Estimate the potential energy of the system in eV, taking the zero of potential energy at infinite separation of electron from from proton.
(b) What is the minimum work required to free the electron, given that its KE in the in the orbit is half the magnitude of potential energy obtained in (a) ?
(c) What are the answers to (a) and (b) above, if zero of potential energy is taken at 1.06Å separation?
Work required to free the electron = 13.58 eV
(c) Potential energy at a separation of r1 (=1.06Å) is
Potential energy of the system, when zero of P.E. is taken at r1 = 1.06Å is
= P.E. at r1 – P.E. at r = 13.58 – 27.16 = -13.58eV
By shifting the zero of potential energy, work required to free the electron is not affected. It
continues to be the same, being equal to + 13.58 eV
q2, q3 = charge on two protons, each = 1.6 × 10-19
r12 = distance between q1 and q2 = 1Å = 10-10 m
r23 = distance between q2 and q3 = 1.5 Å = 1.5× 10-10
r31 = distance between q3 and q1 = 1 Å = 10-10 m.
Taking zero of potential energy at infinity, we have
become equal. After sharing, the charges on two spheres would be
A sharp and pointed end can be treated of very small radius and a flat portion leaves as a sphere of
much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much
higher than on its flatter portions.
(i) What is the electrostatic potential at the points (0,0, z) and (x,y,0) ?
(ii) Obtain the dependence of potential on the distance r of the point from the origin, when
(iii) How much work is done in moving a small test charge from the point (5,0,0)to (-7,0,0) along the x-axis. Does the answer change if path of test charge between the same point is not along the x-axis ?
(iv) If the above point charges are now placed in the same positions in a uniform external electric field what would ne the potential energy of the charge system in its orientation of unstable equilibrium ? Justify your answer in each case.
As work done = charge (V2-V1) ∴ w = zero
As work done by electrostatic field is independent of the path connecting the two points, therefore,
work done will continue to be zero along every path.
– 2q and + q at A,B and C respectively.
Let AC = 2a, we have to calculate electric potential at any point p where BP = r using superposition
principle, potential at P is given by
[You will realise from your answer why ordinary capacitors are in the range of μ F or less. However, electrolytic capacitors do have a much larger capacitance (0-1 F) because of very minute separation between the conductors].
In as 100 pF and 200 pF capacitors are in series, pot. Diff. across C4 is in the ratio 2 : 1.
(a) How much electrostatic energy is stored by the capacitor ?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume (u). Hence arrive at a relation between U and the magnitude of electric field E between the plates.
The origin of factor 1/2 in force can be explained by the fact that inside the conductor, field is zero
and outside the conductor, the fiend is E. Therefore, the average value of the field ( i. e. E/2)
contributes to the force.
C=(4∈_(0 r_1 r_2 ))/(r_1 r_2 ),
where r1 and r2 are the radii of outer and inner spheres respectively.
r1 The induced charge – Q spreads uniformly on the outer surface of inner sphere of radius r2 . the
outer surface of outer sphere is earthed. Due to electrostatic shielding E = 0 for r < r2 and E = 0, for r> r1.
In the space between the two spheres., electric intensity E exists as shown. Potential difference
between the two spheres.
(a) Determine the capacitance of the capacitor. (b) What is the potential of the inner sphere ?
(c) Compare the capacitance of the capacitor with that of radius 12 cm. Explain why the latter is much smaller ?
The capacity of an isolated sphere is much smaller because in a capacitor, outer sphere is earthed,
potential difference decreases and capacitance increases.
(b) If coulomb law involved 1/r3 dependence (instead of 1/r2), would Gauss’s law be still true?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the line of force passing through that point ?
(d) What is the work done by the field of a nucleus in a complete circular orbit of electron ? What if the orbit is elliptical ?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there ?
(f) What meaning would you give to the capacity of a single conductor ?
(g) Guess a possible reason why water has a much greater dielectric constant K = 80 than, say mica (K=6).
non- uniform. Therefore, coulomb’s law is not valid. Hence the magnitude of force is not given
exactly by this formula.
(b) No, Gauss’s law will not be true if coulomb’s law involved 1/r3 dependence instead of 1/r2
(c) The line of force gives the direction of acceleration of charge. If the electric line of force is linear,
the test charge will move along the line. If the line of force is non linear, the charge will not go along
(d) force due to the field is directed towards the nucleus, and the electron does not move in the
direction of this force, therefore work done is zero when the orbit is circular. This is true even when
orbit is ellipitical as electric forces are conservative forces.
(e) No, electric potential is continuous.
(f) The capacity a single conductor implies that the second conductor is at infinity.
(g) This is because a molecule of water in its normal has an unsymmetrical shape and, therefore it
has a permanent dipole moment.
(a) a constant electric field in the Z direction
(b) a fiend that uniformly increases in magnitude but remains in a constant, say z direction
(c) a single positive charge at the origin.
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
four cases given above :
(a) Equipotential surfaces are planes parallel to X – Y plane. These are equidistant.
(b) Equipotential surfaces are planes parallel to X – Y plane. As the field increases uniformly, distance
between the planes (differing by fixed potential) decreases.
(c)Equipotential surfaces are concentric spheres with origin at the Centre.
(d) Equipotential surfaces have the shape which changes periodically. At far off distances from the
grid, the shape of equipotential surfaces becomes parallel to the grid itself.
[ You will learn from this exercise why one cannot build an electrostatic generator using a very small shell, which requires a small charge to acquire a high potential.]
Max. electric field, E = 10% dielectric strength
Obviously. We cannot build an electrostatic generator, using a very small shell.
are connected by a wire, charge will flow essentially from the sphere to the sell. Whatever be the
magnitude and sign of charge q2.
(b) A man fixes outside his house one evening a two Metra high insulating slab carrying on its top, a large aluminium sheet of area 1 m2. Will he get an electric shock If he ouches the metal sheet next morning ?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral ? In other words, what keeps the atmosphere is dissipated ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning ?
ground from an equipotential surface. As we step out into the open from our house, the original
equipotential surfaces of open air change, keeping our body and the ground at the same potential.
That is why we do not get an electric shock.
(b) Yes, the man will get a shock. This is because the steady discharging current of the atmosphere
charges up the aluminium sheet gradually and raises its voltage to an extent depending on the
capacitance of the condenser formed by the aluminium sheet, the ground and the insulating slab.(c)
The atmosphere is being charged continuously by thunderstorms and lightning all over the globe. It
is also discharging due to the small conductivity of air. The two opposing processes, on an average,
are in equilibrium. Therefore, the atmosphere keeps charged.
(d) During a lightning, the electrical energy of the atmosphere is dissipated in the form of light, heat