When an electric dipole is placed in a uniform electric field, it experiences torque and tends to align it in such a way to attain stable equilibrium. Small amount of work done in rotating the dipole through a small angle d \(\theta\) against the torque is given by
dW = \(\tau\) d \(\theta\) = pE sin \(\theta\) d \(\theta\)
\(\therefore\) Total work done in rotating the dipole from orientation \(\theta\)₁ to \(\theta\)₂, \( W=\int_{\theta _{1}}^{\theta_{2}}\)pE sin \(\theta\) d \(\theta\)
= pE(cos\(\theta\)₁ - cos\(\theta\)₂)
\(\Rightarrow\) W = pE (cos\(\theta\)₁ - cos \(\theta\)₂)
Similarly, potential energy of electric dipole, when it rotates
from \(\theta\)₁ = 90° to \(\theta\)₂ = \(\theta\),
W = pE(cos90° - cos\(\theta\)) = -pEcos\(\theta\) = - p.E
W = -p . E

Let the electrostatic potential be zero at point P between the two charges separated by a distance x metre.

At point P,   V_p = V₁ + V₂ = 0
\(\Rightarrow \quad \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{r_{1}}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{r_{2}}=0\)
\(\Rightarrow \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 \times 10^{-6}}{x}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\left(-2 \times 10^{-6}\right)}{(1-x)}=0\)
\(\Rightarrow \quad \frac{4 \times 10^{-6}}{x}=\frac{2 \times 10^{-6}}{(1-x)}\)
\(\Rightarrow \quad \frac{4}{x}=\frac{2}{(1-x)}\)
\(\Rightarrow\)  2 (1-x)=  x
\(\Rightarrow \ 2=3 x \ \text { or } \ x=\frac{2}{3}\)
\(\therefore\) Electrostatic potential is zero at a distance 2/3 m from charge 4\(\mu\)C between the two charges.

Given, p = 4 \(\times\)10^-9 C-m, E = 5 \(\times\) 10⁴,
\(\theta\) = 30°
\(\therefore\) \(\tau\) = pE sin \(\theta\)
= 4 \(\times\)10^-9 \(\times\)5 \(\times\)10⁴ \(\times\)sin 30°
\(=4\times 10^{-9}\times 5\times 10^{4}\times \frac{1}{2}\)    [\(\because\)sin 30° =\(\frac{1}{2}\)]
= 10 \(\times\)10^-5 = 10^-4 N-m

(i) At point A, E =\(\frac { \sigma }{ 2{ \varepsilon }_{ 0 } } =\frac { 2\times{ 10 }^{ 17 } }{ 2\times8.85\times{ 10 }^{ -12 } } \)
= 1.12 x 10²⁸ N/C    
(ii) At point Y. Because at some, the charge sheet acts as a finite sheet acts thus the magnitude remains same towards the middle region of the plane sheet

ABCDEF is a regular hexagon of side 10 cm each. At each corner, the charge q =5 \(\mu\)C is placed. O is the centre of the hexagon.

Given, AB = BC = CD = DE
= EF = FA = d = 10 cm
As, the hexagon has six equilateral triangles, so the distance of centre O from every vertex is 10 cm.
i.e. OA = OB = OC = OD
= OE = OF = d = 10 cm
\(\therefore\) Potential at point O = Sum of potentials at centre O due to individual point charge
i.e. V_O = V_A + V_B + V_C + V_D + V_E + V_F
\(=\frac{1}{4\pi \varepsilon_{0}}.\left [ \frac{q}{OA}+\frac{q}{OB}+\frac{q}{OC}+\frac{q}{OD}+\frac{q}{OE}+\frac{q}{OF} \right ]\)
\(=\frac{1}{4\pi \varepsilon _{0}}.\frac{6q}{d}\)    \(\left [ \because V=\frac{1}{4\pi \varepsilon _{0}.\frac{q}{r}} \right ]\)
Putting the values, we get
\(=9\times 10^{9}\times \frac{6\times 5 \times 10^{-6}}{10\times 10^{-2}}\)
= 2.7 \(\times\)10⁶ V

(i) Here, p = 6.2 x10^-30 C-m and E = 10⁶ N/C
\(\therefore\) \(\tau\) = pE sin\(\theta\)     [for maximum value  \(\theta\)= 90°]
= pE sin 90° = 6.2 x 10^-30 x 10⁶ x 1
= 6.2 x 10^-24 N-m
(ii) When dipole is aligned anti-parallel to the field, \(\theta\) = 180^o
\(\therefore\) W = pE(1 - cos\(\theta\) )
= 6.2 x 10^-30 x 10⁶(1- cos180^o) [\(\because\) cos 180°= -1]
= 6.2 x 10^-30 x 10⁶(1 -(-1))
= 6.2 x 10^-30 x 10⁶ x 2
= 1.24 x 10^-23 J

Electric flux, Φ = −1.0 x 10³ N m²/C
Radius of the Gaussian surface,
r = 10.0 cm
Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −10³ N m²/C.
(b) Electric flux is given by the relation,
\(\phi=\frac{q}{\epsilon_{0}}\)
Where,
q = Net charge enclosed by the spherical surface
∈₀ = Permittivity of free space = 8.854 x 10⁻¹² N⁻¹C² m^{−2
\(\therefore q=\phi \in_{0}\)}
= −1.0 x 10³ x 8.854 x 10⁻¹²
= −8.854 x 10⁻⁹ C
= −8.854 nC
Therefore, the value of the point charge is −8.854 nC.

(a) We can see from the figure that on the left face E and ΔS are parallel. Therefore, the outward flux is
φ_L= E.ΔS = – 200 \(\hat{i}\). ΔS
= + 200 ΔS, since\(\hat{i}\) . ΔS = – ΔS
= + 200 x \(\pi \) (0.05)² = + 1.57 N m² C^–1
On the right face, E and ΔS are parallel and therefore φ_R = E.ΔS = + 1.57 N m² C^–1
(b) For any point on the side of the cylinder E is perpendicular to ΔS and hence E.ΔS = 0. Therefore, the flux out of the side of the cylinder is zero
(c) Net outward flux through the cylinder
φ = 1.57 + 1.57 + 0 = 3.14 N m² C^–1

(d) The net charge within the cylinder can be found by using Gauss’s law which gives
q = ε₀φ
= 3.14 x 8.854 × 10^–12 C
= 2.78 x 10^–11 C

Here, \(q=-1.6\times 10^{-19}C\)
\(dr=5 cm=5\times 10^{-2}m, dV=3000V, m=?\)
\(E=-{dV\over dr}={-3000\over 5\times 10^{-2}}=-6\times 10^4 Vm^{-1}\)
As the charged particle remains suspended in equilibrium, therefore
F = mg = qE
or   \(m={qE\over g}={-1.6\times 10^{-19}\times (-6\times 10^4)\over 9.8}\)
\(=9.8\times 10^{-16}kg\)

(1) Given, C₁ = 2pF, C₂ = 3pF and C₃ = 4pF.
Equivalent capacitance for the parallel combination is given by C_eq .
Therefore, C_eq = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9pF
Hence, the total capacitance of the combination is 9pF.
(2) Supply voltage, V = 100V
The three capacitors are having the same voltage, V = 100v
q = VC
where,
q = charge
C = capacitance of the capacitor
V = potential difference
for capacitance, c = 2pF
q = 100 x 2 = 200pC = 2 x 10^-10C
for capacitance, c = 3pF
q = 100 x 3 = 300pC = 3 x 10^-10C
for capacitance, c = 4pF
q = 100 x 4 = 400pC = 4 x 10^-10 C

(1) Given,
Radius of spherical conductor, r = 12cm = 0.12m
Charge is distributed uniformly over the surface, q = 1.6 x 10^-7 C.
The electric field inside a spherical conductor is zero.
(2) Electric field E, just outside the conductor is given by the relation
\(\mathrm{E}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{r^{2}}\)
Here, permittivity of free space and \(\frac{1}{4 \pi \epsilon_{o}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\)
Therefore,
\(\mathrm{E}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}}=10^{5} \mathrm{NC}^{-1}\)
Therefore, just outside the sphere the electric field is 4.4 x 10⁴ NC^-1.
(3) From the centre of the sphere the electric field at a point 18m = E₁.
From the centre of the sphere, the distance of point d = 18 cm = 0.18m
\(\mathrm{E}_{1}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{d^{2}}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{\left(1.8 \times 10^{-2}\right)^{2}}=4.4 \times 10^{4} \mathrm{NC}^{-1}\)
So, from the centre of sphere the electric field at a point 18 cm away is 4.4 x 10⁴ NC^-1.

Work done in moving a charge q along a closed rectangular path ABCD is calculated as
\( W=W_{A B}+W_{B C}+W_{C D}+W_{D A} W=q E+0-q E+0=0[\because A B=C D] \)

The minimum charge that is stable, is charge of an electron. Since electrons are transferred from one object to another, therefore, electric charge is said to be quantised.

(1) An equipotential surface is defined as the surface over which the total potential is zero. In the given question the plane is normal to line AB. The plane is located at the mid – point of the line AB as the magnitude of the charges are same.
(2) At every point on this surface the direction of the electric field is normal to the plane in the direction of AB.

Here, q = 01 \(\mu\)C = 10^-7C , 2l = 2.0 cm = 2 x 10^-2m,
E = 10⁵ N/C  \(\Rightarrow\) \(\tau\) = pEsin \(\theta\) = q x 2l x E sin\(\theta\) 
\(\therefore\) \(\tau\)_max = 10^-7x 2 x 10 ^-2 x 10⁵ x 1   [\(\because\) sin \(\theta\)  90^o =1]
= 2 x 10^-4 N-m  

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.
The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Energy is stored in the form of electrostatic potential energy in the electric field between the plates of capacitors.