h = 1 x 10^-3 m, e = 1.6 x 10^-19 C, m_e = 9.1 x 10^-31 kg, m_p =1.6 x 10^-27 kg
(a) (i) Time of flight for electron \(t_{c}=\sqrt{\frac{2 h m_{e}}{e E}}\)
Where,
\(E =\frac{V}{d}=\frac{5}{1 \times 10^{-3}}=5000 \mathrm{Vm}^{-1} \)
\(t_{e} =\sqrt{\frac{2 \times 1 \times 10^{-3} \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 5000}}=\sqrt{\frac{18.2 \times 10^{-34}}{8 \times 10^{-16}}}=\sqrt{2.275 \times 10^{-18}} \)
= 1.5 x 10^-9s
t_e = 1.5 ns (ignoring the gravity)
(ii) Time of flight for proton \(t_{p}=\sqrt{\frac{2 h m_{p}}{e E}}\)
\(t_{p} =\sqrt{\frac{2 \times 1 \times 10^{-3} \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times 5000}} \)
\(=\sqrt{\frac{3.2 \times 10^{-30}}{8 \times 10^{-16}}}=\sqrt{\frac{3.2}{8} \times 10^{-14}}=\sqrt{4000 \times 10^{-18}} \)
t_p = 63 x 10^-9s
t_p = 63 ns (ignoring the gravity)
(b) Time of flight for neutron, \(t_{n}=\sqrt{\frac{2 h}{g}}\)
\(t_{n} =\sqrt{\frac{2 \times 1 \times 10^{-3}}{10}}=\sqrt{2 \times 10^{-4}} \)
\(=1,414 \times 10^{-2} \)
\(=14.14 \times 10^{-3} \mathrm{~s} \)
t_n = 14.14 ms
(c) The value of Time period for electron is the least. So, electron will reach the bottom first.

Figure (a)

Parallel : \(C_{P}=C_{0}+C_{0}=2 C_{0}\)
Series : \(\frac{1}{C_{S}} =\frac{1}{C_{0}}+\frac{1}{2 C_{0}} \)
\(=\frac{2+1}{2 C_{0}} \)
\(\frac{1}{C_{S}} =\frac{3}{2} C_{0} \)
\(\therefore C_{s} =\frac{2}{3} C_{0} \)
Figure (b)

\(C_{p 1} =2 C_{0},C_{p 2} =2 C_{0}\)
\(\frac{1}{C_{S}} =\frac{1}{C_{p_{1}}}+\frac{1}{C_{P 2}}=\frac{1}{2 C_{0}}+\frac{1}{2 C_{0}} \)
\(\frac{1+1}{2 C_{0}} =\frac{2}{2 C_{0}}=\frac{1}{C_{0}} \)
\(\therefore C_{s} =C_{0} \)
Figure (c)

\(C_{p} =C_{0}+C_{0}=2 C_{0} \)
\(C_{p} =2 C_{0}+C_{0} \)
\(C_p=3 C_{0} \)
Figure (d)

\(\frac{1}{C_{s 1}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{C_{1}+C_{2}}{C_{1} C_{2}} \)
\(\therefore C_{S 1}=\frac{C_{1} C_{2}}{C_{1}+C_{2}} \)   ....(1)
\(\frac{1}{C_{s 2}}=\frac{1}{C_{3}}+\frac{1}{C_{4}}=\frac{C_{4}+C_{3}}{C_{3} C_{4}} \)
\(\therefore C_{s 2}=\frac{C_{3} C_{4}}{C_{3}+C_{4}} \)    ...(2)
\(C_{p} =C_{s 1}+C_{s 2} \)
\(C_p= \frac{C_{1} C_{2}}{C_{1}+C_{2}}+\frac{C_{3} C_{4}}{C_{3}+C_{4}} \)
\(=\frac{C_{1} C_{2}\left(C_{3}+C_{4}\right)+C_{3} C_{4}\left(C_{1}+C_{2}\right)}{\left(C_{1}+C_{2}\right)\left(C_{3}+C_{4}\right)} \)
\(C_{p} =\frac{\mathbf{C}_{1} C_{2} C_{3}+C_{1} C_{2} C_{4}+C_{3} \mathbf{C}_{4} C_{1}+C_{3} C_{4} C_{2}}{\left(C_{1}+C_{2}\right)\left(C_{3}+C_{4}\right)} \)
The Effective capacitance across PQ and RS is same.
\(C_{PQ}=C_{RS}=C=\frac{C_{1} C_{2} C_{3}+C_{2} C_{3} C_{4}+C_{1} C_{2} C_{4}+C_{1} C_{3} C_{4}}{\left(C_{1}+C_{2}\right)\left(C_{3}+C_{4}\right)}\)
Figure (e): Across PQ

\(\frac{1}{C_{S 1}}=\frac{1}{C_{0}}+\frac{1}{C_{0}}=\frac{2}{C_{0}} \)
\(\therefore C_{S 1}=\frac{C_{0}}{2} \)
\(\frac{1}{C_{S 2}}=\frac{1}{C_{0}}+\frac{1}{C_{0}}=\frac{2}{C_{0}} \)
\(C_{S 2}=\frac{C_{0}}{2} \)
\(C_{P}=C_{S 1}+C_{S 2}=\frac{C_{0}}{2}+\frac{C_{0}}{2} \)
\(C_p=C_{0} \)
Resultant Capacitance

C_R = C_o + C_o
C_R = 2C_o

Distance of separation d = 0.6 x 10^-3 m
Electric field E = 3 x 10⁶ Vm^-1
\(E=\frac{V}{d}\)
(a)  Potential difference V = E x d
V = 3 x 10⁶ x 0.6 x 10^-3
V = 1800 V
(b) If the gap (distance d) is increased then potential difference would be increased, since V = E x d
(c) If E = 3 x 10⁶ Vm^-1 and distance of separation d = 1 x 10^-3 m
Potential difference V = E x d
V = 3 x 10⁶ x 10^-3
V = 3 x 10³ = 3000 V

(i) Electric field is given by \(|E|=\frac{d V}{d x}\)
(a) For region A,
dV = 8 - 5 = 3V, dx - (0.2 - 0) = 0.2m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{3}{0.2}=15 \mathrm{Vm}^{-1}\)
(b) For region B,
dv = 0, E_x = 0
(c) For region C
dV = (7 - 5) = 2V, dx = (0.6 - 0.4) = 0.2 m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{2}{0.2}=10 \mathrm{Vm}^{-1}\)
(d) For region D
dV = (7 - 1)=6V,  dx = (0.8 - 0.6) = 0.2 m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{6}{0.2}=30 \mathrm{Vm}^{-1}\)
(ii) Electric field as a function of x for Figure (ii).

G = 6.67 x 10^-11 Nm²kg^-2
Mass of Earth m_E = 5.9 x 10²⁴ kg
Mass of Moon m_m = 7.9 x 10²² kg
Gravitational force \(F_{g}=\frac{G m_{E} \times m_{M}}{r^{2}}\); Electro static force \(F_e=k\frac{q \times q}{r^2}\)
By equating the forces, \( k\frac{q \times q}{r^{2}}=G \cdot \frac{m_{E} \times m_{M}}{r^{2}} \)
\(\because k=\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \)
\( 4 \pi \varepsilon_{0} =0.11 \times 10^{-9} \)
\(q =\sqrt{4 \pi \varepsilon_{0} G m_{E} \cdot m_{M}} \)   ....(1)
\(=\sqrt{0.11 \times 10^{-9} \times 6.67 \times 10^{-11} \times 5.9 \times 10^{24} \times 7.9 \times 10^{22}}\)
\( q =\sqrt{34.2 \times 10^{26}} \)
q ≈ 5.85 x 10¹³ C
b) Suppose the distance (r) between Moon and Earth is halved, there is no change in the value of charge (q). Because from equation (1), q is independent of distance (r).

From the figure F₁ = F₅ but opposite in direction. So cancel each other
Also, F₂ sin 45^o = F₄ sin 45^o but opposite in direction. So cancel each other.

\( \mathrm{F}_{\mathrm{tot}} =\mathrm{F}_{3}+\mathrm{F}_{2} \cos 45^{\circ}+\mathrm{F}_{4} \cos 45^{\circ} \)
\(\mathrm{F}_{\mathrm{tot}} =\frac{k q Q}{R^{2}}+\frac{k q Q}{R^{2}} \frac{1}{\sqrt{2}}+\frac{k q Q}{R^{2}} \frac{1}{\sqrt{2}} \)
\(\mathrm{~F}_{\mathrm{tot}} =\frac{k q Q}{R^{2}}+2 \frac{k q Q}{R^{2}} \frac{1}{\sqrt{2}} \)
\(=\frac{k q Q}{R^{2}}+\sqrt{2} \sqrt{2} \frac{k q Q}{R^{2}} \frac{1}{\sqrt{2}} \)
\(\vec{F}_{\text {tot }} =\frac{k q Q}{R^{2}}[1+\sqrt{2}] \hat{i} N\)
\(\vec{F} =\frac{1}{4 \pi \varepsilon_{o}} \frac{q Q}{R^{2}}(1+\sqrt{2}) \hat{i N}\)

(a) The electrostatic potential on the surface of the sphere A is V_A = \(\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1 } } \)
The electrostatic potential on the surface of the sphere A is V_B = \(\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 2 } }{ { r }_{ 2 } } \)
Since V_A = V_B. We have
\(\frac { { q }_{ 1 } }{ { r }_{ 1 } } =\frac { { q }_{ 2 } }{ { r }_{ 2 } } \Rightarrow { q }_{ 1 }=\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) { q }_{ 2 }\)
But from the conservation of total charge, Q = q₁ + q₂, we get q₁ = Q – q₂. By substituting this in the above equation,
Q - q₂ = \(\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) { q }_{ 2 }\)
so that q₂ = Q\(\left( \frac { { r }_{ 2 } }{ { r }_{ 1 }+{ r }_{ 2 } } \right) \)
Therefore,
q₂ = 100 x 10^-9 x \(\left( \frac { 2 }{ 10 } \right) \) = 20nC and q₁ = Q - q₂ = 80nC
The electric charge density for sphere A is σ₁ = \(\frac { { q }_{ 1 } }{ 4\pi { r }_{ 1 }^{ 2 } } \)
The electric charge density for sphere B is σ₂ = \(\frac { { q }_{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } \)
Therefore,
σ₁ = \(\frac { 80\times 10^{ -9 } }{ 4 \pi \times 64\times 10^{ -4 } } \) = 0.99 x 10^-6 Cm^-2 and
σ₂ =\(\frac { 20\times 10^{ -9 } }{ 4\pi \times 4\times 10^{ -4 } } \)  = 3.9 x 10^-6 Cm^-2
Note that the surface charge density is greater on the smaller sphere compared to the larger sphere (σ₂ ≈ 4σ₁) which confirms the result \(\frac { { \sigma }_{ 1 } }{ \sigma _{ 2 } } =\frac { { r }_{ 2 } }{ { r }_{ 1 } } \)
The potential on both spheres is the same. So we can calculate the potential on any one of the spheres
V_A=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1 } } =\frac { 9\times 10^{ 9 }\times 80\times { 10 }^{ -9 } }{ 8\times 10^{ -2 } } \)  = 9kV

The capacitors 1 μF and 3μF are connected in parallel and 6μF and 2 μF are also separately connected in parallel. So these parallel combinations reduced to equivalent single capacitances in their respective positions, as shown in the figure (b).
C_eq = 1 + 3 = 4μF
C_eq = 6 + 2 = 8μF
From the figure (b), we infer that the two 4 μF capacitors are connected in series and the two 8 μF capacitors are connected in series. By using formula for the series, we can reduce to their equivalent capacitances as shown in figure (c).
\(\frac { 1 }{ { C }_{ eq } } =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \)  ⇒ C_eq = 2μF and
\(\frac { 1 }{ C_{ eq } } =\frac { 1 }{ 8 } +\frac { 1 }{ 8 } =\frac { 1 }{ 4 } \)  ⇒ C_eq = 4μF 
From the figure (c), we infer that 2μF and 4μF are connected in parallel. So the equivalent capacitance is given in the figure (d).
C_eq = 2 + 4 = 6μF
Thus the combination of capacitances in figure (a) can be replaced by a single capacitance 6 μF.

(a) The capacitance of the capacitor in the presence of dielectric is
C = \(\frac { { \varepsilon }_{ r }{ \varepsilon }_{ 0 }A }{ d } =\frac { 5\times 8.85\times 10^{ -12 }\times 6 \times 10^{-4}}{ 6\times 10^{ -3 } } \)
= 44.25 x 10^-13F = 4.425 pF
The stored charge is
Q = CV = 44.25 x 10^-13 x 10
= 442.5 x 10^-13C = 44.25pC
The stored energy is
\(U=\frac { 1 }{ 2 } \) CV² = \(\frac { 1 }{ 2 } \) x 44.25 x 10^-13 x 100
= 2.21 x 10^-10 J
(b) After the removal of the dielectric, since the battery is already disconnected the total charge will not change. But the potential difference between the plates increases. As a result, the capacitance is decreased.
New capacitance is
C₀=\(\frac { C }{ { \varepsilon }_{ r } } =\frac { 44.25\times 10^{ -12 } }{ 5 } \)
= 0.885 x 10^-12 F = 0.885 pF
The stored charge remains same and 44.25 pC. Hence newly stored energy is
U₀ =\(\frac { { Q }^{ 2 } }{ 2{ C }_{ 0 } } =\frac { { Q }^{ 2 }{ \varepsilon }_{ r } }{ 2C } =\varepsilon _{ r }U\)
= 5 x 2.21 x 10^-10J = 11.05 x 10^-10J
The increased energy is ΔU = (11.05 - 2.21) x 10^-10 J  = 8.84 x 10^-10 J
When the dielectric is removed, it experiences an inward pulling force due to the plates. To remove the dielectric, an external agency has to do work on the dielectric which is stored as additional energy. This is the source for the extra energy 8.84 x 10^–10 J.

(a) The capacitance of the capacitor is
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } =\frac { 8.85\times { 10 }^{ -12 }\times 25\times 10^{ -4 } }{ 1\times 10^{ -3 } } \)
= 221.2 x 10^-13 F
C = 22.12 x 10^-12F = 22.12pF
(b) The charge stored in any one of the plates is Q = CV, Then
Q = 22.12 x 10^-12 x 10 = 221.2 x 10^-12C = 221.2 pC.

If the conductor has no net charge, then its motion is the same as usual projectile motion of a mass m which we studied in Kinematics (unit 2, vol-1 XI physics). Here, in this problem, in addition to downward gravitational force, the charge also will experience a downward uniform electrostatic force.
The acceleration of the charged ball due to gravity = -g\(\hat { j } \)
The acceleration of the charged ball due to uniform electric field =\(-\frac { qE }{ m } \hat { j } \)
The total acceleration of charged ball in downward direction \(\vec { a } =-\left( g+\frac { qE }{ m } \right) \vec { j } \)
It is important here to note that the acceleration depends on the mass of the object. Galileo’s conclusion that all objects fall at the same rate towards the Earth is true only in a uniform gravitational field. When a uniform electric field is included, the acceleration of a charged object depends on both mass and charge.
But still the acceleration a = \(\left( g+\frac { qE }{ m } \right) \) is constant throughout the motion. Hence we use kinematic equations to calculate the range, maximum height and time of flight. In fact we can simply replace g by \(g+\frac { qE }{ m } \) in the usual expressions of range, maximum height and time of flight of a projectile.

	Without charge	With the charge +q
Time of flight T	\(\frac { 2v_{ 0 }sin\theta }{ g } \)	\(\frac { 2{ v }_{ 0 }sin\theta }{ \left( g+\frac { qE }{ m } \right) } \)
Maximum height hmax	\(\frac { { v }_{ 0 }^{ 2 }sin^{ 2 }\theta }{ 2g } \)	\(\frac { { v }_{ 0 }^{ 2 }sin^{ 2 }\theta }{ 2\left( g+\frac { qE }{ m } \right) } \)
Range R	\(\frac { { v }_{ 0 }^{ 2 }sin2\theta }{ g } \)	\(\frac { { v }_{ 0 }^{ 2 }sin2\theta }{ \left( g+\frac { qE }{ m } \right) } \)

Note that the time of flight, maximum height, range are all inversely proportional to the acceleration of the object. Since \(\left( g+\frac { qE }{ m } \right) \) >g for charge +q, the quantities T, h_max, and R will decrease when compared to the motion of an object of mass m and zero net charge. Suppose the charge is –q, then \(\left( g-\frac { qE }{ m } \right) \)< g, and the quantities T, h_max and  R will increase. Interestingly the trajectory is still parabolic as shown in the figure.

(a) Electric potential at point P is given by
V_p=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }_{ p } } =\frac { 9\times 10^{ 9 }\times 9\times { 10 }^{ -6 } }{ 10 } \)  = 8.1 x 10³ V
Electric potential at point Q is given by
V_Q = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }_{ Q } } =\frac { 9\times 10^{ 9 }\times 9\times { 10 }^{ -6 } }{ 16 } \) = 5.06 x 10³ V
Note that the electric potential at point Q is less than the electric potential at point P. If we put a positive charge at P, it moves from P to Q. However if we place a negative charge at P it will move towards the charge +9μC.
The potential difference between the points P and Q is given by
ΔV = V_p-V_Q = +3.04 x 10³ V
b) Suppose we replace the charge +9 μC by -9 μC, then the corresponding potentials at the points P and Q are,
V_p = -8.1 x 10³ V, V_Q = -5.06 x 10³ V
Note that in this case electric potential at the point Q is higher than at point P.
The potential difference or voltage between the points P and Q is given by
ΔV = V_p-V_Q= -3.04 x 10³V
(c) The electric potential V at a point P due to some charge is defined as the work done by an external force to bring a unit positive charge from infinity to P. So to bring the q amount of charge from infinity to the point P, work done is given as follows.
W = qV
W_Q = 2 x 10^-6 x 5.06 x 10³J = 10.12 x 10^-3 J.

Case (a)
The magnitude of the electric field at point P is
E_p = \(\frac { 1 }{ 4\pi \varepsilon _{ 0 } } \frac { q }{ { r }^{ 2 } } =\frac { 9\times { 10 }^{ 9 }\times 1\times 10^{ -6 } }{ 4 } \)
= 2.25 x 10³ NC^-1
Since the source charge is positive, the electric field points away from the charge. So the electric field at the point P is given by
\(\bar { { E }_{ p } } \) = 2.25 x 10³ NC^-1
For the point Q
\(|\vec { { E }_{ Q } } |=\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ 16 } \) = 0.56 x 10³ NC^-1
Hence \(\vec { { E }_{ Q } } \) = 0.56 x 10³\(\hat { j } \) NC^-1
Case (b)
The magnitude of the electric field at point P
\(\bar { { E }_{ p } } =\frac { kq }{ r^{ 2 } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } =\frac { 9\times 10^{ 9 }\times 2\times 10^{ -6 } }{ 4 } \)
= 4.5 x 10³ NC^-1
Since the source charge is negative, the electric field points towards the charge. So the electric field at the point P is given by
\(\vec { E_{ p } } \) = -4.5 x 10³\(\hat { i } \)NC^-1
For the point Q, \(|\vec { { E }_{ Q } } |\frac { 9\times 10^{ 9 }\times 2\times { 10 }^{ -6 } }{ 36 } \)
= 0.5 x 10³ NC^-1
\(\vec { E_{ Q } } \) = 0.5 x 10³\(\hat { i } \)NC^-1
At the point Q the electric field is directed along the positive x-axis.

According to the superposition principle, the total electrostatic force on charge q₁ is the vector sum of the forces due to the other charges,
\(\vec { { F }_{ 1 }^{ tot } } =\bar { { F }_{ 12 } } +\bar { { F }_{ 13 } } +\bar { F_{ 14 } } \)
The following diagram shows the direction of each force on the charge q₁.

The charges q₂ and q₄ are equi-distant from q₁. As a result the strengths (magnitude) of the forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \) are the same even though their directions are different. Therefore the vectors representing these two forces are drawn with equal lengths. But the charge q₃ is located farther compared to q₂ and q₄. Since the strength of the electrostatic force decreases as distance increases, the strength of the force \(\vec { { F }_{ 13 } } \) is lesser than that of forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \). Hence the vector representing the force \(\vec { { F }_{ 13 } } \) is drawn with smaller length compared to that for forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \).
From the figure, r₂₁ =\(\sqrt { 2 } \) m = r₄₁ and r₃₁ = 2m
The magnitudes of the forces are given by
F₁₃ = \(\frac { kq^{ 2 } }{ r_{ 31 }^{ 2 } } =\frac { 9\times 10^{ 9 }\times 10^{ -12 } }{ 4 } \)
F₁₃  = 2.25 x 10^-3 N
F₁₂ = \(\frac { kq^{ 2 } }{ r_{ 31 }^{ 2 } } ={ F }_{ 14 }=\frac { 9\times 10^{ 9 }\times 10^{ -12 } }{ 2 } \)
= 4.5 x 10^-3N
From the figure, the angle θ = 45⁰. In terms of the components, we have
\(\vec { { F }_{ 12 } } ={ F }_{ 12 }cos\theta \hat { i } -{ F }_{ 12 }sin\theta \hat { j } \)
= 4.5 x 10^-3 x \(\frac { 1 }{ \sqrt { 2 } } \hat { i-4.5\times { 10 }^{ -3 }\times \frac { 1 }{ \sqrt { 2 } } \hat { j } } \)
 \(\vec { { F }_{ 13 } } =F_{ 13 }\hat { i } \) = 2.25 x 10^-3 N\(\hat { i } \)
\(\vec { { F }_{ 14 } } ={ F }_{ 14 }cos\theta \hat { i } +{ F }_{ 14 }sin\theta \hat { j } \)
= 4.5 x 10^-3 x \(\frac { 1 }{ \sqrt { 2 } } \hat { i+4.5\times { 10 }^{ -3 }\times \frac { 1 }{ \sqrt { 2 } } \hat { j } } \)
Then the total force on q₁ is,
\(\vec { { F }_{ 1 }^{ tot } }={ (F }_{ 12 }cos\theta \hat { i } -{ F }_{ 12 }sin\theta \hat { j } )+{ F }_{ 13 }\hat { i } +{ (F }_{ 14 }cos\theta \hat { i } +{ F }_{ 14 }sin\theta \hat { j } )\)
\(\vec { { F }_{ 1 }^{ tot } } =({ F }_{ 12 }cos\theta +F_{13}+{ F }_{ 14 }cos\theta )\hat { i } +(-{ F }_{ 12 }sin\theta +{ F }_{ 14 }sin\theta )\)\(\hat { j } \)
Since F₁₂ = F₁₄, the j ^th component is zero.
Hence we have
\(\vec { { F }_{ 1 }^{ tot } } =({ F }_{ 12 }cos\theta +F_{13}+{ F }_{ 14 }cos\theta )\hat { i } \)
substituting the values in the above equation,
\(\left( \frac { 4.5 }{ \sqrt { 2 } } +2.25+\frac { 4.5 }{ \sqrt { 2 } } \right) \times10^{-3}\hat { i }=(4.5\sqrt { 2 } +2.25)\times 10^{-3}\hat { i } \)
\(\vec { { F }_{ 1 }^{ tot } } \) = 8.61 x 10^-3 N\(\hat { i } \)
The resultant force is along the positive x-axis.

The proton and the electron attract each other. The magnitude of the electrostatic force between these two particles is given by
\(F_e=\frac { ke^{ 2 } }{ { r }^{ 2 } } =\frac { 9\times 10^{ 9 }\times (1.6\times 10^{ -19 })^{ 2 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
=\(\frac { 9\times 2.56 }{ 28.09 } \) x 10^-7 = 8.2 x 10^-8 N
The gravitational force between the proton and the electron is attractive. The magnitude of the gravitational force between these particles is
F_G = \(\frac { G{ m }_{ e }{ m }_{ p } }{ { r }^{ 2 } } \)
= \(\frac { 6.67\times 10^{ -11 }\times 9.1\times 10^{ -31 }\times 1.6\times 10^{ -27 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
= \(\frac { 97.11 }{ 28.09 } \) x 10^-47 = 3.4 x 10^-47N
The ratio of the two forces \(\frac { { F }_{ e } }{ F_{ G } } =\frac { 8.2\times 10^{ -8 } }{ 3.4\times 10^{ -47 } } \)
= 2.41 x 10³⁹
Note that F_e ≈ 10³⁹ F_G
The electrostatic force between a proton and an electron is enormously greater than the gravitational force between them. Thus the gravitational force is negligible when compared with the electrostatic force in many situations such as for small size objects and in the atomic domain. This is the reason why a charged comb attracts an uncharged piece of paper with greater force even though the piece of paper is attracted downward by the Earth. This given figure is shown in below.

Electrostatic attraction between a comb and pieces of papers

If the two spheres are neutral, the angle between them will be 0^o when hanged vertically. Since they are positively charged spheres, there will be a repulsive force between them and they will be at equilibrium with each other at an angle of 30° with the vertical. At equilibrium, each charge experiences zero net force in each direction. We can draw a free-body diagram for one of the charged spheres and apply Newton’s second law for both vertical and horizontal directions.
The free-body diagram is shown below

In the x-direction, the acceleration of the charged sphere is zero.
Using Newton’s second law \((\vec { { F }_{ tot }= } m\vec { a } )\), we have
T sinθ\(\hat { i } \) - F_e\(\hat { i } \) =0
T sinθ = F_e    ......(1)
Here T is the tension acting on the charge due to the string and F_e is the electrostatic force between the two charges.
In the y-direction also, the net acceleration experienced by the charge is zero
Tcosθ\(\hat { j } \) - mg\(\hat { j } \) = 0
Tcosθ = mg   ..(2)
By dividing equation (1) by equation (2),
tanθ = \(\frac { { F }_{ e } }{ mg } \)          .....(3)
Since they are equally charged, the magnitude of the electrostatic force is
\({ F }_{ e }=k\frac { { q }^{ 2 } }{ { r }^{ 2 } } \) where k=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \)
Here r = 2a = 2Lsinθ. By substituting these values in equation (3),
tanθ = k\(\frac { { q }^{ 2 } }{ mg(2Lsin\theta )^{ 2 } } \)       ..........(4)
Rearranging the equation (4) to get q
q = 2 Lsinθ\(\\ \sqrt { \frac { mgtan\theta }{ k } } \)
= 2 x 0.1 x sin 30^o x \(\sqrt { \frac { 10^{ -3 }\times 10\times { tan30 }^{ 0 } }{ 9\times 10^{ 9 } } } \)
q = 8.01 x 10^-8C = 80.1 nC

(a) q₁ = +2 μC, q₂ = +3 μC, and r = 1m. Both are positive charges. so the force will be repulsive.
Force experienced by the charge q₂ due to q₁ is given by
\(\vec { { F }_{ 21 } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } \)
Here \(\hat { r_{ 12 } } \) is the unit vector from q₁ to q₂. Since q₂ is located on the right of q₁, we have
\(\hat { r_{ 12 } } =\hat { i } \), and \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } }=9 \times10^9\) so that
\(\vec { { F }_{ 21 } } =\frac { 9\times 10^{ 9 }\times 2\times 10^{ -6 }\times 3\times 10^{ -6 } }{ 1^{ 2 } } \hat { i } \)
= 54 x 10^-3 N\(\hat { i } \) 
According to Newton’s third law, the force experienced by the charge q₁ due to q₂ is \(\vec { { F }_{ 12 } } =-\vec { { F }_{ 21 } } \) Therefore,
\(\vec { F_{ 12 } } \)= 54 x 10^-3 N\(\hat { i } \)
The directions of \(\vec { { F }_{ 21 } } \) and \(\vec { { F }_{ 12 } } \)  are shown in the figure (case (b)).
(b) q₁ = +2 μC, q₂ = –3 μC, and r = 1m. They are unlike charges. So the force will be attractive.
Force experienced by the charge q₂ due to q₁ is given by
\( \vec{F}_{21} =\frac{9 \times 10^{9} \times\left(2 \times 10^{-6}\right) \times\left(-3 \times 10^{-6}\right)}{1^{2}} \hat{r}_{12} \)
\(=-54 \times 10^{-3} \mathrm{~N} \hat{i}\left(\mathrm{Using} \hat{r}_{12}=\hat{i}\right)\)
The charge q₂ will experience an attractive force towards q₁ which is in the negative x direction.
According to Newton’s third law, the force experienced by the charge q₁ due to q₂ is \(\vec{F}_{12}=-\vec{F}_{21}\) Therefore,
\(\vec{F}_{12}=54 \times 10^{-3} \widehat{i} \mathrm{~N}\)
The directions of \(\vec{F}_{21} \text { and } \vec{F}_{12}\) are shown in the figure (case (b)).
(c) If these two charges are kept inside the water, then the force experienced by q₂ due to q₁
\(\vec { { F }_{ 21 }^{ W } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } \)
since ε = ε_rε₀,
we have \(\vec { { F }_{ 21 }^{ W } } =\frac { 1 }{ 4\pi { { \varepsilon }_{ r }\varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \hat { r_{ 12 } } =\frac { \vec { F_{ 21 } } }{ { \varepsilon }_{ r } } \)
Therefore,
\(\vec { { F }_{ 21 }^{ W } } =\frac { 54\times { 10 }^{ -3 }N }{ 80 } \hat { i } \) = -0.675 x 10^-3 N\(\hat { i } \)

  Coulomb force Gravitational force (i)  It is attractive or repulsive. It is always attractive . (ii) It depends on the nature of the medium in
which the two charges are kept at rest. It is independent of the medium. (iii)  It obeys Newton's third law. It does not obey Newton's Third Iaw. (iv) Coulomb force is given by,
\(\overrightarrow{F_{12}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \hat{r}_{21}\) Gravitational force is given by,
\(\vec{F}=G \frac{m_{1} m_{2}}{r^{2}}\hat{r}\)