(a) E/2 (b) E/4
(c) E/6 (d) None of these
ans: Correct option is (B)
If the new distance = 2r
and electric field due to single charge E is inversely proportional r square
therefore new intensity is E/4.
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Q2. The surface density on the copper sphere is σ . the electric field strength on the surface of copper sphere is ?
(a) σ (b) σ/2
(c) σ/2ε0 (d) Q/ε0
ans: Correct option is ((d)
According to Gauss's theorem
E = [ q/4Ï€R²]／ε0
or E = σ/ε0
Q3. In a region of space having a uniform electric field E, a homospherical bowl of a radius r is placed. The electric flux Ф through the bowl is ?
(a) 2Ï€RE (b) 4Ï€R²E
(c) 2Ï€R²E (d) Ï€R²E
ans: Correct option is (c)
Ф = E (ds) cosӨ
= E (2Ï€R²) cos0Ù
= 2Ï€R²E.
Q4. An electric dipole when placed in uniform electric field E will have minimum potential energy if
the dipole moment makes the following angle with E
(a)π (b) π/2
(c) zero (d) 3Ï€/2
ans: Correct optio is (C)
Up = - p . E
= -p E cosÓ¨
= Up (minimum) = -pE
when Ó¨ = 0Ù
Solution:
(a) The charge of one electron is -1.6 × 10-19
C
So net
charge on a substance consisting of 5 × 1014 electrons is
5 × 1014 × (-1.6 × 10-19 C ) = -8 × 10-5 C = -80 µC.
(b)
Similarly the net charge on a substance consisting of a combination of 7
× 1013 Protons and 14 × 1013 electrons is
[ 7 × 1013
× (1.6 × 10-19 C )]+ [ 4 × 1013(-1.6 × 10-19 C
)]
= +4.8 µC.
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Q6. When a piece of polythene is rubbed with
wool , a charge of 2 × 10-7 C is developed on polythene. What is the amount of
mass , which is transferred to polythene?
Solution:
No. of
electrons transferred n = q/e
Mass transferred = me × n = me × ( q/e )
= 9.1 ×
10-31 × ( 2×10-7/1.6 × 10-19)
= 11.38 × 10-19 kg.
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Q7. Four identical point charges each of
magnitude q are placed at the corner of the square of side a. Find the net
electrostatic force on any of the charge.
Solution: Let the concerned charge be at C then charge at C will experience the force due to charges at A, B and D. let this forces respectively be FA, FB and FD thus forces are given as
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Q8. An electron of mass Me , initially at rest, moves through a certain distance in a uniform electric field in time t1. A proton of mass Mp also initially at rest, takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, find the ratio t2/t1.
Solution:
Q9. Can a metal sphere of radius 1cm hold a charge of 1 coulomb.
Solution: Electric field at the surface of the sphere.
Q10. Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be…
(e being the charge of electron)
Solution: Let n be the number of electrons missing.
with the horizontal side of the square as
shown in figure. The electric flux linked to the surface, in units of volt. m,
is… 
Solution:
Q12. When
a body is charged by the induction, then the body
(a) Becomes neutral
(b) Does not lose any charge
(c) Loses whole of the charge on it
(d) Loses part of the charge on it
Solution:
Correct option is (b)
Charging
by induction involves transfer o charges from one part to the another of the
body. No lose any charge is involved.
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Q13. On charging by conduction, mass of a body may
(a) Increase
(b) Decrease
(c) Both (a) & (b)
(d) None of the above
Solution:
correct option is (c)
On
charging by conduction, body may gain mass, if it acquires negative charge. It
may lose mass, if it acquires positive charge.
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Q14. If a body is positively charged, then it has
(a) Excess of electrons
(b) Excess of protons
(c) Deficiency of electrons
(d) Deficiency of neutrons
Solution:
Correct option is (c)
Positive
charge is due to deficiency of electrons.
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Q15. Shown in the figure are two positive charges +Q and –Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If σ1 is the surface charge on the inner surface and Q1 net charge on it and σ2 the surface on the outer surface and Q2 net charge on it then:
(a) σ1 ≠ 0, Q1 = 0 σ2 = 0, Q2 = 0
(a) σ1 ≠ 0, Q1 = 0 σ2 ≠ 0, Q2 = 0
(c) σ1 = 0, Q1 = 0 σ2 = 0, Q2 = 0
(d) σ1 ≠ 0, Q1 ≠ 0 σ2 ≠ 0, Q2 ≠ 0
Solution
: Correct option is (c)
Inside
the cavity net charge is zero.
Q1 = 0
and σ1 = 0
There
is no effect of point charges +Q, -Q and induced charge on inner surface on the
outer surface .
σ2 = 0 and Q2 = 0
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