The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external electric field \(\vec{E}.\)
 \(V=\frac{1}{4\piε_0}\frac{q}{r}\)

(a) \(C_{1}=\frac{\varepsilon_{0} A}{2d} (\because area=\frac{A}{2})\)
\(C_{2}=\frac{\varepsilon_{r} \varepsilon_{0} A}{d} \)
C₁ and C₂ are in parellel,
\(C_{P}=\left(C_{1}+C_{2}\right) \)
\(C_{P}=\frac{1}{2}\left(\frac{\varepsilon_{r} \varepsilon_{0} A}{d}+\frac{\varepsilon_{0} A}{d}\right) \)
\(=\frac{1}{2} \frac{\varepsilon_{0} A}{d}\left(\varepsilon_{r}+1\right) \)
\(C_{P}=\frac{\varepsilon_{0} A}{2 d}\left(\varepsilon_{r}+1\right) \)

(b) \(C_1=\frac{\varepsilon_0 \varepsilon_r A}{d / 2}=\frac{2 \varepsilon_0 \varepsilon_r A}{d}(\because \text { Separation }=d / 2)\) 
\(C_2=\frac{2 \varepsilon_0 A}{d}\)
C₁ and C₂ are in series,
\(\frac{1}{C_1}  =\frac{1}{C_1}+\frac{1}{C_2}  \)
\(=\frac{d}{2 \varepsilon_0 \varepsilon_r A}+\frac{d}{2 \varepsilon_0 A}  \)
\(=\frac{d}{2 \varepsilon_0 A}\left(\frac{1}{\varepsilon_r}+1\right)  \)
\(\frac{1}{C_1}  =\frac{d}{2 \varepsilon_0 A}\left(\frac{1+\varepsilon_r}{\varepsilon_r}\right)  \)
\(\therefore C_s  =\frac{2 e_0 A}{d}\left(\frac{e_r}{1+e_r}\right)\)

C_p = C_b + C_c
\(C_{P}=6+2=8 \mu \mathrm{F} \)
\(\frac{1}{C_{s}}=\frac{1}{C_a}+\frac{1}{C_p}=\frac{1}{C_d} \)
\(C_{s}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8} \)
\(\therefore C_{s}=\frac{8}{3} \mu \mathrm{F} \)
Total capacitance \(C_{s}=\frac{8}{3} \times 10^{-6} \mathrm{~F} \)
Total Charge, \(Q=C_{s} V=\frac{8}{3} \times 10^{-6} \times 9 \)
\(Q_{a}=24 \mu C\)
(a) Charge on capacity \(Q_{a}=24 \mu C\)         .....(1)
Charge on capacitor \(Q_{b}=24 \times \frac{6}{8}=18 \mu \mathrm{C} \)             ..............(2)
Charge on capacitor  \(Q_{c}=24 \times \frac{2}{8}=6 \mu \mathrm{C} \)              ..............(3)
Charge on capacitor \(Q_{d}=24 \times \frac{8}{8}=24 \mu \mathrm{C} \)              ..............(4)
(b) Potential difference across \(C_{a} \ is\ V_{a}=\frac{Q_{a}}{C_{a}} \)
\(=\frac{24}{8}=3 \mathrm{~V} \)      ...(5)
Potential difference across \(C_{b}\ is \ V_{b}=\frac{Q_{b}}{C_{b}} \)
\(=\frac{18}{6}=3 \mathbf{V}\)       ........(6)
Potential difference across \(C_{c}\ is \ V_{c}=\frac{Q_{c}}{C_{c}}=\frac{6}{2}=3 \mathrm{~V} \)       ......(7)
Potential difference across \(C_{d}\ is \ V_{d}=\frac{Q_{d}}{C_{d}} \)
\(=\frac{24}{8}=3 \mathbf{V} \)          ....(8)
(c) Energy stored in each capacitor \(U=\frac{1}{2} C V^{2}\)
Energy stored in \(\mathrm{C}_{\mathrm{a}} \text { is } U_{a}=\frac{1}{2} C_{a} V_{a}^{2}\)
\(U_{\mathrm{a}}=\frac{1}{2} \times 8 \times 10^{-6} \times 3 \times 3=36 \mu \mathrm{J}\)   ....(9)
Energy stored in \(C_{b}\ is \ U_{b}=\frac{1}{2} C_{b} V_{b}^{2} \)
\(U=\frac{1}{2} \times 6 \times 10^{-6} \times 3 \times 3 \)
\(=27 \mu \mathrm{J} \)         ........(10)
Energy stored in C_e is \(U_{c} =\frac{1}{2} C_{c} V_{c}^{2} \)
\(=\frac{1}{2} \times 2 \times 10^{-6} \times 3 \times 3=9 \mu \mathrm{J} \)          ......(11)
Energy stored in C_d is \(U_{d} =\frac{1}{2} C_{d} V_{d}^{2} \)
\(U_d=\frac{1}{2} \times 8 \times 10^{-6} \times 3 \times 3 \)
\(=36 \times 10^{-6} \mathrm{~J}=36 \mu \mathrm{J} \)      .........(12)

(a) Electric Field E = \(\frac{Potential \ difference}{Distance}=\frac{V}{d}\)
Electric field E = 3 x 10⁶ Vm^-1
Distance d = 1000 m
∴ Potential difference V = E x d =3 x 10⁶ x 10³ = 3 x 10⁹ V
(b) Potential Energy U = qV
U = 25 x 3 x 10⁹ = 75 x 10⁹ J
∴ Potential energy transferred to the ground = 75 x 10⁹ J

\(W =\left(V_{b}-V_{a}\right) q\left[where\ V=\frac{K Q}{r}\right] \)
To find V_b :
\(V_b=\frac{kQ}{r_3}+\frac{kQ}{r_4}\)
\(V_{b} =\frac{K \times 10 \times 10^{-6}}{\sqrt{50 \times 10^{-4}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{250 \times 10^{-4}}} \)
\(V_{b} =\frac{K \times 10^{-5}}{\sqrt{50} \times 10^{-2}}+\frac{K \times 10^{-5}}{\sqrt{250} \times 10^{-2}} \)
\(=K \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
\(=9 \times 10^{9} \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
V_b = 1842002 V
To find V_a :
\(V_a=\frac{kQ}{r_1}+\frac{kQ}{r_2}\)
\(V_{a} =\frac{K \times 10 \times 10^{-6}}{\sqrt{5 \times 10^{-2}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{15 \times 10^{-2}}} \)
V_a = 2400000 V
\(\therefore W_{D} =\left(V_{b}-V_{a}\right) q=(1842002-2400000)( -2 \times 10^{-6} )\)
W = +1.12J
Positive sign implies that to move the charge - 2 μC external work is required.

 E = 2 x 10³ N C^-1
Φ = EA cos θ
a) Vertical rectangular surface : 
\(\phi_{\text {vertical surface }}=E A \cos 180^ {o}\)
\( \phi_{v s}=\left(-2 \times 10^{3} \right) \times(0.05 \times 0.15)\)
\(\phi_{v s}=-15 \mathrm{Nm}^{2} / \mathrm{C} \)
b) Slanted Surface :
\( \phi_{\text {slanted surface }} =\mathrm{EA} \cos 60^{\circ} \)
Here, cos 60^o = 1/2
\(\text { hyp } =\frac{5 \mathrm{~cm}}{\sin 30^{\circ}}=\frac{5}{1 / 2}=10 \mathrm{~cm}=0.1 \mathrm{~m} \)
\( \phi_{\text {slanted surface }}=\frac{1}{2}EA,\)
\(\phi_{s s} =\frac{1}{2}\left(2 \times 10^{3}\right) \times(0.1 \times 0.15) \)
\(\phi_{\mathrm{as}} =15 \mathrm{Nm}^{2} / \mathrm{C} \)
(c) Entire Surface :
\( \phi_{\text {entire surface }}=\phi_{v s}+\phi_{\text {slanted }}+\phi_{\text {ends }} \)
\(\phi_{\text {end sufface }}=\mathrm{EA} \cos 90^{\circ}0\)
\(\phi_{\text {entire surface }}=-15+15 +0 \)
\(\phi_{\text {entire surface }}=0 \)

(a) Acceleration of the electron \(a=\frac{F}{m}\)
Electrostatic force F = - eE
\(a=\frac{F}{m}=\frac{-e E \hat{j}}{m}\)   ... ( 1)
(b) Let velocity of the electron be \(\vec{v}\)
v = u + at        ...(2)
Here, \( u=v_{0} \hat{i} \), \(a=\frac{-e \bar{E}}{m} \hat{j}\)
Substituting these values in the equation (2) we get
\(\therefore \vec{v}=\nu_{0} \hat{i}-\frac{e E}{m} t \hat{j}\)
(c) Displacement \(\vec{r}\) represents position,
\( s=u t+\frac{1}{2} a t^{2} \) ...(3)
Let, \(s=\hat{r} \) and Here, \(u=v_{0} \hat{i} \), \(a=\frac{-e E}{m} \hat{j}\)
Substituting these values in the equation (3), we get,
\(\hat{r}=v_{0} t \hat{i}-\frac{1}{2} \frac{e E}{m} t^{2} \hat{j}\)

(a) In the figure

1 - Electrostatic force F_e = QE
2 - Weight W = mg
3 - Elastic force F = -kx
4 - Upward force = Normal reaction = N
b) In this figure

1- Electrostatic force F = qE
2 - Weight W = mg
3 - Tension acting along the string is T
(c) The charge is attracted towards the positively charged plate because it is a negative charge.

1 - Force = qE
2 - Downward force F = mg

Distance of separation r = 1 m
Number of electron in human body = 10²⁸
Charge appeared on my friend and me, q = 1% of charge on 10²⁸ electrons
q \(=10^{28} \times 1 / 100 \)
q = 1.6 x 10⁷ C
Force between us \( \mathrm{F}_{\mathrm{e}} =\mathrm{Kq}^{2} / \mathrm{r}^{2} \)
\(\left.=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{7}\right)^{2}}{(1)^{2}} \quad \text { (as } \mathrm{r}=1 \mathrm{~m}\right)\)
= 9 x 2.56 x 10⁹ x 10¹⁴
F_e = 23.04 x 10²³ N
Mass of each person m₁ = m₂ = m = 60 kg
Weight of each person W = mg = 60 x 9.8
W = 588N
F_e / W = \(\frac{23.04\times10^{23}}{588} = 3.9 \times 10^{21}=(or)F_e= 3.9 \times 10^{21}W\)

Charge q = 50 nC = 50 x 10^-9 C , e = 1.6 x 10^-19 C
No. of electrons \(n=\frac{q}{e}\)
\(\frac { q }{ e } =\frac { 50\times { 10 }^{ -9 } }{ 1.6\times { 10 }^{ -19 } } \)
=  31.25 x 10¹⁰ electrons
To produce 50 nC charge, number of electrons transferred = 31.25 x 10¹⁰ electrons

When an irregular shaped conductor is given positive charge, the electric field near the sharp end is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called corona discharge.

The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors. \(C=\frac{Q}{V}\)
Its unit is Coulomb per volt or farad (F).

i) When the external electric field applied to the dielectric is very large, the bound charges (electrons) become free charges. This is called dielectric breakdown.
ii) The maximum electric field the dielectric can withstand before it breaksdown is called dielectric strength.

The alignment of the dipole moments of the permanent or induced dipoles in the direction of applied electric field is called polarisation.

Electrostatic shielding is the process of isolating a certain region of space from external field. It is based on the fact that electric field inside a conductor is zero. This property is called elecrostatic shielding because anything placed inside the cavity of the conductor will be completely shielded from external fields.

The electrostatic energy stored per unit volume of space is called electrostatic energy density.
\(u_E=\frac{1}{2}ε_0E^2\)

The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux.

Electrostatic potential energy is the work done to assemble the charges at the given locations brought from infinity. If q₁ and q₂ are the charges to be assembled with a separation of distance r, the electric potential energy is, \(U=\frac{1}{4\piε_o}\frac{q_1q_2}{r}\)

i) The electric field is the negative gradient of the electric potential. \(E=\frac{-dV}{dx}\). 
ii) In vector form, \(\vec{E}=-\left[\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \hat{k}\right]\)

(i) The work done to move a charge q between any two points A and B, W = q (V_B - V_A). If the points A and B lie on the same equipotential surface, work done is zero because V_B = V_A.
(ii) The electric field is normal to an equipotential surface. 

An equipotential surface is a surface on which all the points are at the same electric potential.

For a collection of n point charges, the electric dipole moment is defined as follows, \(\vec{p}=\stackrel{i=n} \sum _{i=1}q_i\vec{r}_i\) where,  \(\vec{r}_i\) is the position vector of charge q_i from the origin.

(i) Two equal and opposite charges separated by a small distance constitute an electric dipole.
(ii) The magnitude of the electric dipole moment is equal to the product of the magnitude of one of the charges and the distance between them, \(|\vec{p}|=2 q a\).
(iii) The electric dipole moment vector lies along the line joining two charges and is directed from -q to +q.

(i) No two electric field lines intersect each other. lf two lines cross at a point, then there will be two different electric field vectors at the same point.
(ii) As a consequence, if some charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence electric field lines do not intersect.

Electric field lines are a set of continuous lines which represent the electric field in some region of space visually.

Electric field at the point P at a distance r from the point charge q is the force experienced by a unit positive placed at that point P and is given by
\(\vec { E } =\frac { \vec { F } }{ q_{ 0 } } =\frac { kq }{ { r }^{ 2 } } \hat { r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } \hat { r } \)
where \(\hat{r}\) is the unit vector pointing from q to the point of interest P.

It there are more than two charges, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
Consider a system of n charges namely q₁, q₂, q₃ ...q_n. The force on q₁ exerted by the charge q₂ is \(\overrightarrow{F_{12}}=k \frac{q_{1} q_{2}}{r_{21}^{2}} \hat{r}_{21}\) .
The force on q₁ exerted by the charge q₃ is \(\overrightarrow{F_{13}}=k \frac{q_{1} q_{3}}{r_{31}^{2}} \hat{r}_{31}\)
By continuing this, the total force acting on the charge q₁ due to all other charges is given by
\( \vec{F}_{1}^{\text { tot }}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\overrightarrow{F_{14}}+\ldots+\vec{F}_{1 n} \)
\(\vec{F}_{1} ^{\text { tot }}=k\left\{\frac{q_{1} q_{2}}{r_{21}^{2}} \hat{r}_{21}+\frac{q_{1} q_{3}}{r_{31}^{2}} \hat{r}_{31}+\frac{q_{1} q_{4}}{r_{41}^{2}} \hat{r}_{41}+\ldots+\frac{q_{1} q_{n}}{r_{n 1}^{2}} \hat{r}_{n 1}\right\}\) 

Coulomb's law \(\overrightarrow{F_{21}}=\frac{k q_{1} q_{2}}{r^{2}} \hat{r}_{12}\)
where, q₁ - charge; q₂ - charge
r - distance between the charges
\(\hat{r}_{12}\)- the unit vector directed from charge q₁ to charge q₂
k = Proportionality constant

The charge of an electron is the elementary charge in nature. Therefore, charge on any body is the integral multiple of an electron. The charge on any body can be expressed by the formula,
q = ne; where, n is the number of electrons, e is the charge on one electron.
n = 0, ±1, ±2, ±3, ±4, ...
This is called quantization of charge.

The electric field on the surface of the sphere(by Gauss law) is given by
E = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q }{ { R }^{ 2 } } \)
The potential on the surface of the hollow metallic sphere is given by
V = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q }{ { R } } \) = ER
Since V_max = E_maxR
Here E_max = 3 x 10⁶ Vm^-1. So the maximum potential difference created is given by
V_max = 3 x 10⁶ x 0.5
= 1.5 x 10⁶V (or) 1.5 million volt.

(i) In figure (a), area A₁ encloses the charge Q. So electric flux through this closed surface A₁ is \(\frac { Q }{ { \varepsilon }_{ 0 } } \). But the closed surface A₂ contains no charges inside, so electric flux through A₂ is zero.
(ii) In figure (b), the net charge inside the cube is 3q and the total electric flux in the cube is therefore \(\Phi _{ E }=\frac { 3q }{ { \varepsilon }_{ 0 } } \).
Note that the charge -10 q lies outside the cube and it will not contribute the total flux through the surface of the cube.

The electric flux through the rectangular area
\({ \Phi }_{ E }=\vec { E } .\vec { A } \) = EA cosθ = 100 x 5 x 10 x 10^-4 x cos60°
⇒ \({ \Phi }_{ E }\) = 0.25 Nm²C^-1
For θ = 0°
\({ \Phi }_{ E }=\vec { E } .\vec { A } \)= EA
= 100 x 5 x 10 x 10^-4 = 0.5 Nm²C^-1

When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) - U(0°)
From the equation U =−pE cosθ = −\(\hat p.\hat E\) ,
we write U = − pE cosθ, Next, we calculate the work done to rotate one water molecule from θ = 0° to 90°.
For one water molecule
W = - pE cos90^o + pE cos0^o  = pE
W=  6.3 x 10^-30 x 3 x 10⁵ = 18.9 x 10^-25J
For 10²² water molecules, the total work done is
W_tot = 18.9 x 10^-25 x 10²² = 18.9 x 10^-3J

(a) The work done to arrange the charges in the corners of the square is independent of the way they are arranged. We can follow any order.
(i) First, the charge +q is brought to the corner P. This requires no work since no charge is already present, W_P = 0
(ii) Work required to bring the charge –q to the corner Q = (-q) x potential at a point Q due to +q located at a point P
W_Q = -q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ a } =-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^{ 2 } }{ a } \)
(iii) Work required to bring the charge +q to the corner R = q x potential at the point R due to charges at the point P and Q.
W_R = q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( -\frac { q }{ a } +\frac { q }{ \sqrt { 2 } a } \right) \)
= \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }^{ 2 } }{ a } \left( -1+\frac { 1 }{ \sqrt { 2 } } \right) \)
(iv) Work required to bring the fourth charge –q at the position S = q × potential at the point S due the all the three charges at the point P, Q and R.
W_s = - q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ a } +\frac { q }{ a } -\frac { q }{ \sqrt { 2 } a } \right) \)
W_s = \(-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^2 }{ a } \left( 2-\frac { 1 }{ \sqrt { 2 } } \right) \)
(b) Work required to bring the charge q′ to the center of the square = q′ x potential at the center point O due to all the four charges in the four corners.
The potential created by the two +q charges are canceled by the potential created by the -q charges which are located in the opposite corners. Therefore the net electric potential at the center O due to all the charges in the corners is zero.
Hence no work is required to bring any charge to the point O. Physically this implies that if any charge q′ when brought close to O, then it moves to the point O without any external force.

In the given problem, since the potential depends only on x, we can use \(\vec { E } =\frac { dV }{ dx } \hat { i } \) (the other two terms \(\frac { \eth V }{ \eth y } \) and \(\frac { \eth V }{ \eth z } \) are zero)
From 0 to 1 cm, the slope is constant and so \(\frac { dV }{ dx } \) = 25V cm^-1, So \(\vec { E } \)= -25V cm^-1\(\hat { i } \)
From 1 to 4 cm, the potential is constant
V = 25 V. It implies that \(\frac { dV }{ dx } \) = 0, So \(\vec { E } \) = 0
From 4 to 5 cm, the slope \(\frac { dV }{ dx } \) = -25 V cm^-1
So \(\vec { E } \) = + 25Vcm^-1\(\hat { i } \)
The plot of electric field for the various points along the x-axis is given below

The maximum torque experienced by the dipole is when it is aligned perpendicular to the applied field.
ፒ_max = pE sin 90⁰ = 3.4 x 10^-30 x 3 x 10⁴ Nm
ፒ_max = 10.2 x 10^-26 Nm.

Case (a) The position vector for the +q on the positive x-axis is a\(\hat { i } \) and position vector for the +q charge the negative x-axis is - a\(\hat { i } \) So the dipole moment is,
\( \vec { p } =(+q)(a\hat { i } )+(+q)(-a\hat { i } )=0\)
Case (b) In this case one charge is placed at the origin, so its position vector is zero. Hence only the second charge +q with position vector a\(\hat { i } \) contributes to the dipole moment, which is \(\vec { p } =qa\hat { i } \).
From both cases (a) and (b), we can infer that in general the electric dipole moment depends on the choice of the origin and charge configuration. But for one special case, the electric dipole moment is independent of the origin. If the total charge is zero, then the electric dipole moment will be the same irrespective of the choice of the origin. It is because of this reason that the electric dipole moment of an electric dipole (total charge is zero) is always directed from –q to +q, independent of the choice of the origin.
Case (c) \(\vec { p } =(-2q)a\hat { j } +q(2a)(-\hat { j } )=-4qa\hat { j } \) 
Note that in this case \(\vec { p } \) is directed from -2q to +q.
Case (d) \(\vec { p } =-2qa(-\hat { i } )+qa\hat { j } +qa(-\hat { j } )\)
= \(2qa\hat { i } \).
The water molecule (H₂O) has this charge configuration. The water molecule has three atoms (two H atom and one O atom). The centers of positive (H) and negative (O) charges of a water molecule lie at different points, hence it possess permanent dipole moment. The electric dipole moment  \(\vec { p } \) is directed from center of negative charge to the center of positive charge, as shown in the figure.

(i) The electric field lines start at q₂ and end at q₁. In figure (a), q₂ is positive and q₁ is negative. The number of lines starting from q₂ is 18 and number of the lines ending at q₁ is 6. So q₂ has greater magnitude. The ratio of \(\left| \frac { { q }_{ 1 } }{ { q }_{ 2 } } \right| =\frac { { N }_{ 1 } }{ { N }_{ 2 } } =\frac { 6 }{ 18 } =\frac { 1 }{ 3 } \). It implies that |q₂| = 3|q₁|.
(ii) In figure (b), the number of field lines emanating from both positive charges are equal (N = 18). So the charges are equal. At point A, the electric field lines are denser compared to the lines at point B. So the electric field at point A is greater in magnitude compared to the field at point B. Further, no electric field line passes through C, which implies that the resultant electric field at C due to these two charges is zero.
(iii) In the figure (c), the electric field lines start at q₁ and q₃ and end at q₂. This implies that q₁ and q₃ are positive charges. The ratio of the number of field lines is \(\left| \frac { { q }_{ 1 } }{ { q }_{ 2 } } \right| =\frac { 8 }{ 16 } =\left| \frac { { q }_{ 3 } }{ { q }_{ 2 } } \right| =\frac { 1 }{ 2 } \), implying that q₁ and q₃ are half of the magnitude of q₂. So q₁ = q₃ = +10 nc.

Note: A similar problem is solved in XIth Physics volume I, unit 3 section 3.3.2. There are three forces that acts on the mass m:
(i) The downward gravitational force exerted by the Earth (mg)
(ii) The normal force exerted by the inclined surface (N)
(iii) The Coulomb force given by uniform electric field (qE) The free body diagram for the mass m is drawn below.

A convenient inertial coordinate system is located in the inclined surface as shown in the figure. The mass m has zero net acceleration both in x and y-direction.
Along x-direction, applying Newton’s second law, we have
mg sinθ\(\hat { i } \) - qE\(\hat { i } \) = 0
mg sinθ - q E = 0
or, E = \(\\ \frac { mgsin\theta }{ q } \)
Note that the magnitude of the electric field is directly proportional to the mass m and inversely proportional to the charge q. It implies that, if the mass is increased by keeping the charge constant, then a strong electric field is required to stop the object from sliding. If the charge is increased by keeping the mass constant, then a weak electric field is sufficient to stop the mass from sliding down the plane.
The electric field also can be expressed in terms of height and the length of the inclined surface of the plane.
E = \(\frac { mgh }{ qL } \).

By using superposition principle, the net electric field at point A is
\(\vec { E_{ A } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1A }^{ 2 } } \hat { { r }_{ 1A } } +\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 2A }^{ 2 } } \hat { { r }_{ 2A } } \)
where r_1A and r_2A are the distances of point A from the two charges respectively
\(\vec { E_{ A } } =\frac { 9\times 10^{ 4 }\times 1\times { 10 }^{ -6 } }{ (2\times { 10 }^{ -3 })^{ 2 } } (\hat { j } )+\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ (2\times 10^{ -3 })^{ 2 } } (\hat { i } )\)\(\)
= 2.25 x 10⁹\(\hat { j } \) + 2.25 x 10⁹\(\hat { i } \) = 2.25 x 10⁹ (\((\hat { i } +\hat { j } )\)
The magnitude of electric field
\(|\vec { { E }_{ A } } |=\sqrt { (2.25\times { 10 }^{ 9 })^{ 2 }+(2.25\times 10^{ 9 })^{ 2 } } \)
= 2.25 x \(\sqrt { 2 } \) x 10⁹ NC^-1
The direction of \(\vec { { E }_{ A } } \) is given by \(\frac { \vec { { E }_{ A } } }{ |\vec { { E }_{ A } } | } =\frac { 2.25\times 10^{ 9 }(\hat { i } +\hat { j } ) }{ 2.25\times \sqrt { 2 } \times { 10 }^{ 9 } } =\frac { (\hat { i } +\hat { j } ) }{ \sqrt { 2 } } \), which is the unit vector along OA as shown in the figure.

The acceleration experienced by an electron placed at point A is
\(\vec { a_{ A } } =\frac { \vec { F } }{ m } =\frac { q\vec { { E }_{ A } } }{ m } \)
=\(\frac { (-1.6\times 10^{ -19 })\times (2.25\times 10^{ 9 })(\hat { i } +\hat { j } ) }{ 9.1\times { 10 }^{ -31 } } \)
= -3.95 x 10²⁰ \((\hat { i } +\hat { j } )\)Nkg^-1
The electron is accelerated in a direction exactly opposite to \(\vec { { E }_{ A } } \).

According to the quantisation of charge
q = ne
Here q = 1C. So the number of electrons in 1 coulomb of charge is
n = \(\frac { q }{ e } =\frac { 1C }{ 1.6\times 10^{ -19 } } \) = 6.25 x 10¹⁸ electrons