Numerical Examples
Example 1.
A polythene piece rubbed with wool is found to have a negative charge of 3.2 \times 10^{-7} , \text{C} .
Estimate the number of electrons transferred (from which to which?). Is there any transfer of mass from wool to polythene?
Solution.
An electron has a negative charge of 1.6 \times 10^{-19} , \text{C} . Therefore, the number of electrons equivalent to 3.2 \times 10^{-7} , \text{C} of negative charge is
\frac{3.2 \times 10^{-7}}{1.6 \times 10^{-19}} = 2.0 \times 10^{12} .
Since, the polythene piece on being rubbed with wool, acquires a negative charge of 3.2 \times 10^{-7} , \text{C} , it follows that 2.0 \times 10^{12} electrons have been transferred from wool to polythene.
An electron has a mass of 9.1 \times 10^{-31} , \text{kg} . Hence, there is also a small transfer of mass equivalent to
(2.0 \times 10^{12}) \times (9.1 \times 10^{-31} , \text{kg}) \approx 2 \times 10^{-18} , \text{kg} .
Example 2.
Two equal point-charges Q = +\sqrt{2} , \mu \text{C} are placed at each of the two opposite corners of a square, and equal point-charges q at each of the other two corners. What must be the value of q so that the resultant force on Q is zero?
Solution.
Let a (metre) be the side of the square. The length of a diagonal will be \sqrt{2}a .
The force on the positive charge Q at A due to the positive charge Q at C is
F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q^2}{2a^2}
and is directed outwards, as shown. In order to make net force on Q zero,
the resultant of the other two forces F_1 and F_2 on Q due to the charges q and q should be equal and opposite to F .
Clearly, F_1 and F_2 will be directed as shown and for this both q ’s should be negative.
Now,
F_1 = F_2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qq}{a^2}
Since, F_1 and F_2 are at right angles, their resultant is
F_{12} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{\sqrt{2}Qq}{a^2}
For the equilibrium of Q , we must have
F = F_{12}
or \frac{Q^2}{2a^2} = \frac{\sqrt{2}Qq}{a^2}
or q = \frac{Q}{2\sqrt{2}}
But Q = +\sqrt{2} , \mu \text{C} (given).
\therefore \quad q = \frac{\sqrt{2} , \mu \text{C}}{2\sqrt{2}} = -0.5 , \mu \text{C} .
Example 3. Two pieces of copper, each of mass 10 g, are 10 cm apart from each other. One electron per 1000 atoms is transferred from one piece of copper into the other. How much Coulombian force will act between them after the transference of electrons.
Atomic weight of copper is 63.5 g/mol, Avogadro’s number = 6 \times 10^{23} , \text{/mol} , charge on electron = 1.6 \times 10^{-19} , \text{C} and \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 , \text{Nm}^2 \text{C}^{-2} .
Solution. 63.5 g of copper has 6 \times 10^{23} atoms.
Hence, the number of atoms in 10 g of copper is
\frac{6 \times 10^{23}}{63.5} \times 10 .
1 electron is transferred per 1000 atoms of a piece of copper. Hence, the number of electrons transferred from one piece of copper into the other is
\frac{6 \times 10^{23} \times 10}{63.5} \times \frac{1}{1000} = 9.45 \times 10^{19} .
Charge on 1 electron is 1.6 \times 10^{-19} , \text{C} .
Hence, the (negative) charge transferred from one piece into the other is
9.45 \times 10^{19} \times (1.6 \times 10^{-19}) = 15.12 , \text{C} .
An equal amount of positive charge will be left in the first piece. Now, the force of attraction between the two pieces is given by
F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r^2}
Here, r = 10 , \text{cm} = 0.1 , \text{m}, \quad q_1 = q_2 = 15.12 , \text{C}
\Rightarrow F = (9 \times 10^9) \times \frac{(15.12)^2}{(0.1)^2} = 2.057 \times 10^{14} , \text{N} .
Example 4.
Two electric charges q and 2q are at a distance a apart from each other in air. A third charge Q is to be placed along the same line in such a way that the net force acting at q and also at 2q is zero.
Calculate the position of charge Q in terms of q and a .
Solution.
Suppose the charge Q is placed in between the charges q and 2q , at a distance x from the charge q .
Its distance from the charge 2q will be (a - x) . The net force on q will be zero when the force F on it due to 2q is equal (and opposite) to the force F' due to Q . That is,
\frac{1}{4\pi\varepsilon_0} \cdot \frac{q \times 2q}{a^2} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q \times Q}{x^2} \tag{i}
For this, if q and 2q are both positive, then Q should be negative.
Similarly, the net force on 2q will be zero when the force on it due to q is equal (and opposite) to the force due to Q . That is,
\frac{1}{4\pi\varepsilon_0} \cdot \frac{2q \times q}{a^2} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2q \times Q}{(a - x)^2} \tag{ii}
From Eqs. (i) and (ii), we have
\frac{q \times Q}{x^2} = \frac{2q \times Q}{(a - x)^2}
or (a - x)^2 = 2x^2
or a - x = \pm \sqrt{2}x
or x = \frac{a}{1 + \sqrt{2}}
The minus sign is inadmissible because Q is placed in between q and 2q .
The charge Q should be placed at a distance of \frac{a}{(1 + \sqrt{2})} from the charge q.
Example 5. A particle of mass m and charge +q is located midway between two fixed charged particles each having a charge +q and at a distance 2L apart. Assuming that the middle charge moves along the line joining the fixed charges, calculate the frequency of oscillation when it is displaced slightly.
Solution. Suppose the movable charge +q is displaced through a distance x from the middle point C of the line. Then, the forces acting on it due to the fixed charges 1 and 2 are given by:
F_1 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{(L + x)^2} and
F_2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{(L - x)^2}
Resultant force,
F = F_1 - F_2 = \frac{4 q^2 L x}{4 \pi \varepsilon_0 (L^2 - x^2)^2}
For x \ll L ,
F = - \frac{q^2 x}{\pi \varepsilon_0 L^3}
The negative sign indicates that F is a restoring force which tends to bring the charge +q back into the position C. The acceleration of the charge due to this force is given by:
a = \frac{F}{m} = - \frac{q^2 x}{m \pi \varepsilon_0 L^3} = - \omega^2 x , where
\omega^2 = \frac{q^2}{m \pi \varepsilon_0 L^3}
Hence, the motion of the charge is simple harmonic whose frequency of oscillation is given by:
n = \frac{\omega}{2\pi} = \frac{q}{2\pi \sqrt{m \pi \varepsilon_0 L^3}}
Example 6.
Two identical charged spheres each of mass 100 g are suspended in air by two strings each of length 100 cm and make an angle of 2\theta with each other. Due to repulsion, the spheres remain at 5 cm distance from each other. Find the charge on each sphere. What will happen if the system is taken to a gravity free space (say a satellite)?
Solution. Suppose the mass of each sphere is m kg, the charge on each q coulomb and in equilibrium the distance between them is r metre. Each sphere is in equilibrium under the action of three forces:
weight of the sphere = m g
electrical force of repulsion F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}
tension T in the string
Resolving these forces in vertical and horizontal components, we have:
T \cos \theta = m g and
T \sin \theta = F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}
So,
\tan \theta = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{m g r^2} \tag{i}
Since the distance between spheres r \ll length of string ( l )
So that \theta will be very small and we can take
\tan \theta \approx \sin \theta
Then Eq. (i) becomes,
\sin \theta = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{m g r^2}
\frac{r}{2} = \frac{1}{4 \pi \varepsilon_0 m g} \cdot \frac{q^2}{r^2}
or
q^2 = 4 \pi \varepsilon_0 m g \cdot \frac{r^3}{2l}
Given:
m = 100 \times 10^{-3} , \text{kg}, \quad g = 10 , \text{m/s}^2, \quad r = 5 \times 10^{-2} , \text{m}, \quad l = 1.0 , \text{m}
and 4 \pi \varepsilon_0 = \frac{1}{9 \times 10^9}
Thus,
q^2 = \frac{1}{9 \times 10^9} \cdot \frac{100 \times 10^{-3} \times 10}{2 \times 1.0} \cdot \left( \frac{125}{10^6} \right) = \frac{625}{9} \times 10^{-16}
So that,
q = \pm \frac{25}{3} \times 10^{-8} , \text{C}
When the system is taken to a gravity free space, the strings will become straight (at an angle of 180° with each other).
Example 7.
Three point-charges of +2 , \mu\text{C}, -3 , \mu\text{C} , and -3 , \mu\text{C} are kept at the vertices A, B , and C respectively of an equilateral triangle of side 20 cm, as shown.
What should be the sign and magnitude of a charge to be placed at the mid-point M of side BC so that the charge at A remains in equilibrium?
Solution. The forces on the charge at A due to the charges at B and C are equal in magnitude and are given by:
F_1 = F_2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(2 \times 10^{-6})(3 \times 10^{-6})}{(0.2)^2}
These are directed as shown in the figure below. Their resultant is given by:
F_{12} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(2 \times 10^{-6})(3 \times 10^{-6})}{(0.2)^2} \cdot (2 \cos 30^\circ)
= \frac{1}{4\pi\varepsilon_0} \cdot \frac{(6 \times 10^{-12}) \cdot \sqrt{3}}{0.04}
and is directed along AM .
Suppose, a charge +q is placed at M for the equilibrium of the charge at A .
(The charge q should necessarily be positive for balancing the force F_{12} ).
The force on the charge at A due to the charge q is given by:
F' = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(2 \times 10^{-6})q}{(AM)^2}
But
(AM)^2 = (AB)^2 - (BM)^2 = (0.2)^2 - (0.1)^2 = 0.03 , \text{m}^2
So,
F' = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(2 \times 10^{-6})q}{0.03}
The force F' is directed as shown. For the equilibrium of the charge at A , we must have:
F' = F_{12}
\Rightarrow \frac{1}{4\pi\varepsilon_0} \cdot \frac{(2 \times 10^{-6})q}{0.03} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{6 \times 10^{-12} \cdot \sqrt{3}}{0.04}
Solving:
q = \frac{6 \times 10^{-12} \cdot \sqrt{3} \cdot 0.03}{2 \times 10^{-6} \cdot 0.04} = \frac{9\sqrt{3}}{4} , \mu\text{C}