Q.1
2 equal charges, 24 micro Coulomb each, are separated by 5 cm. Find force
between these.
Charge Q1 = 24𝜇𝐶 = 24x10-6 C
Charge Q2 = 24𝜇𝐶 = 24x10-6 C
Distance D = 5 cm = 5x10-2 m
Force between charges
𝐹=
𝐹 = (9 × 10
9)
1 𝑄1 𝑄2
4𝜋𝜖0 𝐷2
(24 × 10−6 ) × (24 × 10−6 )
×
= 2073.60 𝑁
(5 × 10−2 )2
Q.2
2 equal charges, 199 micro Coulombs each, experiences an attraction force of 199
Newton. Calculate the distance between the charges in mm.
Charge Q1 = 199𝜇𝐶 = 199x10-6 C
Charge Q2 = 199𝜇𝐶 = 199x10-6 C
Force between charges F = 199 N
𝐹=
199 = (9 × 10
9)
1 𝑄1 𝑄2
4𝜋𝜖0 𝐷2
(199 × 10−6 ) × (199 × 10−6 )
×
𝐷2
𝐷 = 1.338 𝑚 = 1338 𝑚𝑚
Q.3
Add 0.005 Coulombs, 2,555 micro Coulombs and 363,286 nano Coulombs and
express the result in micro Coulombs.
Charges
𝑄1 = 0.005𝐶 = 0.005 × 106 𝜇𝐶 = 5000 𝜇𝐶
𝑄2 = 2555𝜇𝐶
𝑄3 = 363286 𝑛𝐶 = 363286 × 10−3 𝜇𝐶 = 363.286𝜇𝐶
Total Charge
𝑄 = 5000 + 2555 + 363.286 = 7918.286 𝜇𝐶
Q.4
2 equal charges are separated by 37 mm and experiences a repulsion force of 9
Newton. Calculate the amount of each charge in micro Coulombs.
Charge Q1 = Q2 = Q
Distance D = 37mm = 37x10-3 m
Force between charges F = 9N
𝐹=
9 = (9 × 10
1 𝑄1 𝑄2
4𝜋𝜖0 𝐷2
9)
𝑄2
×
(37 × 10−3 )2
𝑄 = 1.17 × 10−6 𝐶 = 1.17𝐶
Q.5
What is the total charge of all protons in 3 gram of water (H2O)?
Molecular weight of H2O= 2x1+16=18
Moles in 3 gms = 3/18 = 1/6 M
Number of molecules = moles x Avogadro constant =
1.004 × 1023
1
6
× (6.022 × 1023 ) =
Now 1 molecule of H2O has 2 Hydrogen atoms and 1 Oxygen atom.
Hydrogen has 1 proton per atom and Oxygen has 8 protons per atom.
Thus 1 molecule of H2O has total 10 (1x2+8) protons per molecule.
So total number of protons in 3 gms water = (1.004 x 1023) x 10 =1.004 x 1024
Charge on one proton = 1.6x10-19C
So total charge on protons = (1.004 x 1024) x (1.6 x 10-19) = 160640 C
Q.6
How many excess electrons can be counted in a rubber rod of - 1 *0.00001 nano
Coulombs of charges?
Total charge = -1 x 0.00001 nC = -1 x 10-5 nC = (-1 x 10-5) x 10-9 C= -1 x 10-14 C
Charge on one electron = - 1.6 x 10-19C
Number of electrons =
−1×10−14
−1.6×10−19
= 62500
Q.7
A metal ball has - 2 Coulomb of charge. If it receives 0.50*10^20 number of
electrons, what will be the resultant charge of the ball?
Original charge = - 2C
Added electrons = 0.50 x 1020
Charge on one electron = - 1.6 x 10-19C
So newly added charge = (0.50 x 1020) x (-1.6 x 10-19) = - 8C
Net charge = -2 -8 = -10 C
Q.8
How many electrons have to be added to 0 kg metal sphere such that another +1C
charge located 2 mm above the sphere will be able to hold that in the air?[Assume
g=10m/s^2].
Data insufficient as mass of sphere is given 0 Kg
Assuming mass is 𝑚 Kg (known) and charge is Q C (unknown)
Distance = 2 mm = 2 x 10-3 m
Gravitational force 𝐹1 = 𝑚𝑔 = 𝑚 × 103 × 10 = 10000𝑚 𝑁
Attraction force
𝐹2 =
1 𝑄1 𝑄2
1×𝑄
9)
(9
=
×
10
×
= 2.25 × 1015 𝑄 𝑁
(2 × 10−3 )2
4𝜋𝜖0 𝐷2
Under equilibrium
𝐹1 = 𝐹2
2.25 × 1015 𝑄 = 10000𝑚
𝑄 = 4.444 × 10−12 𝑚 𝐶
Number of electrons =
𝑄
1.6×10−19
= 1.7 × 107 𝑚