Net outward flux through the surface of the box, Φ = 8.0 × 10³ N m²/C
For a body containing net charge q, flux is given by the relation,
\(\phi=\frac{q}{\epsilon_{0}}\)
∈₀ = Permittivity of free space
= 8.854 × 10⁻¹² N⁻¹C² m⁻²
q = ∈₀Φ
= 8.854 × 10⁻¹² × 8.0 × 10³
= 7.08 × 10⁻⁸
= 0.07 μC
Therefore, the net charge inside the box is 0.07 μC.
(b) No
Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

R_p = 30 / 7.5 = \(4 \ \Omega\)
\(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
or \(\frac{1}{4}=\frac{1}{10}+\frac{1}{12}+\frac{1}{R_3}\)
or \(\frac{1}{R_3}=\frac{1}{4}-(\frac{1}{10}+\frac{1}{12})=\frac{4}{60}=\frac{1}{15}\)
or R₃ = \(15 \ \Omega\)

Yes, two balls having same kind of charge can attract each other if charge possessed by one ball is very large as compared to that on the other ball because when two charged balls are placed near each other, they induced opposite charges on the faces of each other. On the ball having small amount of charge, very large amount of induced charge is produced, two balls get attracted towards each other.

Electric field intensity at a point is the electric force experienced by a unit positive charge placed at the point.

Its SI unit is NC^-1 or Vm^-1.
\(E_{P}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{2 q}{a^{2}}\right) \Rightarrow \vec{E}_{P}=-\frac{\vec{p}}{4 \pi \varepsilon_{0} a^{3}}\)
It is in the direction opposite to the direction of dipole moment (i.e. from + ve to -ve charge).

Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
E = Ir
\(I=\frac{E}{r}\)
\(=\frac{12}{0.4}=30 A\)
The maximum current drawn from the given battery is 30 A.

Zero

(a) Electric flux through an area is the product of magnitude of area and the component of electric field vector normal to it.
\(\phi_{\mathrm{E}}=\Delta S(E \cos \theta)=\vec{E} \cdot \Delta \vec{S}\)
Its SI unit is NC^-1 m² 
(b) Electric field intensity due to a thin infinite plane sheet charge: Consider a thin infinite sheet of charge with uniform surface charge density a. To calculate electric field at a point P distant r from the sheet we imagine a symmetrical Gaussian surface in such a way that the point charge lies on it. Here we assume a cylinder of cross-sectional area A and length 2r with its axis perpendicular to the sheet.

Flux through the curved surface of the cylinder,
\(\phi_{1}=\int \vec{E} \cdot \overrightarrow{d s}=0 \left(\because \theta=90^{\circ}\right)\)
Total flux through plane faces of the cylinder,
\(\phi_{2}=2 \int \vec{E} \cdot \overrightarrow{d s}=2 E A \left(\because \theta=0^{\circ}\right)\)
Net flux through the Gaussian surface is
\(\phi=\phi_{1}+\phi_{2}=2 E A\)
Net charge enclosed by the Gaussian surface is
\(Q=\sigma A\)
According to the Gauss's theorem, \(\phi=\frac{Q}{\varepsilon_{0}}\)
\(\therefore \phi=\frac{\sigma A}{\varepsilon_{0}}\)
From equations (i) and (ii), we get
\(2 E A=\frac{\sigma A}{\varepsilon_{0}} \Rightarrow E=\frac{\sigma}{2 \varepsilon_{0}}\)
(c) For positively charged sheet, the electric field is directed away from the sheet.
For negatively charged sheet, the electric field is directed towards the plane sheet.

Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns I₁, I₂ and I₃ which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff’s second rule for the closed loop ADCA gives,
10 - 4(I₁ - I₂) + 2(I₂ + I₃ - I₁) - I₁ = 0 
that is, 7I₁ -  6I₂ - 2I₃ = 10
For the closed loop ABCA, we get
10 - 4I₂ - 2 (I₂ + I₃) - I₁ = 0
that is, I₁ + 6I₂ + 2I₃ = 10 
For the closed loop BCDEB, we get
5 - 2 (I₂ + I₃) - 2 (I₂ + I₃ - I₁) = 0
that is, 2I₁ - 4I₂ - 4I₃ = -5
Equations (3.61 a, b, c) are three simultaneous equations in three unknowns. These can be solved by the usual method to give
\(I_1=2.5 \mathrm{~A}, \quad I_2=\frac{5}{8} \mathrm{~A}, \quad I_3=1 \frac{7}{8} \quad \mathrm{~A}\)
The currents in the various branches of the network are 
\(\mathrm{AB}: \frac{5}{8} \mathrm{~A}, \quad \mathrm{CA}: 2 \frac{1}{2} \mathrm{~A}, \quad \mathrm{DEB}: 1 \frac{7}{8} \mathrm{~A}\)
\(\mathrm{AD}: 1 \frac{7}{8} \mathrm{~A}, \quad \mathrm{CD}: 0 \mathrm{~A}, \quad \mathrm{BC}: 2 \frac{1}{2} \mathrm{~A}\)
It is easily verified that Kirchhoff’s second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB
\(5 \mathrm{~V}+\left(\frac{5}{8} \times 4\right) \mathrm{V}-\left(\frac{15}{8} \times 4\right) \mathrm{V}\)
equal to zero, as required by Kirchhoff’s second rule.

Here A = 1m², d = 5 cm, q =?
E = 1000 NC^-1
From \(E={\sigma\over\epsilon_o}={q/A\over \epsilon_o}\)
\(q=A\epsilon_oE=1\times (8.85\times 10^{-12})\times 1000\)
\(=8.85\times 10^{-9}C\)

(i) a
(ii) a
(iii) b
(iv) d
(v) c

(i)

\(\because\) 6\(\Omega\) and 3\(\Omega\) are connected in series
R_eq = 6 + 3
= 9\(\Omega\) 
\(\therefore \ \mathrm{I}=\frac{\mathrm{P} \cdot \mathrm{D} .}{\mathrm{R}_{e q}}\)
\(=\frac{18-0}{9}\)
= 2A
\(\therefore\) V_A - V_a = 6I
18 - 6 x 2 = V _a
\(\therefore\) V_a = 6V                  ..........(i)​​​​​​

6\(\mu\)F and 3\(\mu\)F capacitors are in series.
\(\therefore \ C_{e q}=\frac{6 \times 3}{6+3}\)
= 2\(\mu\) F
\(\therefore\) Charge on 6\(\mu\)F = Charge on 3\(\mu\)F = Charge withdrawn
\(\therefore\) Q = C_eq x P.D.
= 2 x 10^-6 x 18
= 36 x 10^-6 C = 36 \(\mu\)F
\(\therefore\) Potential difference across 6\(\mu\)F is given by
\(18-\mathrm{V}_{b}=\frac{\mathrm{Q}}{6 \times 10^{-6}}\)
\(18-\frac{36 \times 10^{-6}}{6 \times 10^{-6}}=\mathrm{V}_{b}\)
\(\Rightarrow\)V_b = 12V
\(\therefore\) V_ab = 12 - 6 = 6V
(ii) When switch S is closed then
\(\text { (P.D.) }_{6 \Omega}=\text { (P.D.) }_{6 \mu \mathrm{F}} \text { and }(\mathrm{P} . \mathrm{D} .)_{3 \Omega}=(\mathrm{P.D} .)_{3 \mu \mathrm{F}}\) as 6\(\Omega\) resistor and 6\(\mu\)F capacitor are in parallel and, 3\(\Omega\) resistor and 3\(\mu\)F capacitor are in parallel.
\(\therefore\) V_a = V_b = 6V (using eq. (i) in part (a))
(iii) When switch is closed, P.D. across capacitor of capacitance 6\(\mu\)F is = 18 - 6 = 12V
\(\therefore\) Electric charge stored after closing S in it is
Q = CV = 6 x 10^-6 x 12 = 72\(\mu\)C
\(\therefore\) Charge that will flow through capacitor and is equal to = 72 - 36 = 36\(\mu\)C.

(i) (c)
(ii) (b)
(iii) (a): Since, E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface.
(iv) (b): The work done in bringing unit positive charge from infinity to a point which is at a distance x from the positive charge Q is defined as the potential at the given point due to the charge Q. Therefore
\(\phi=W\)
(v) (b): \(W_{\text {ext }}=q_{0} \Delta V\)
\(\left(W_{A B}\right)_{\mathrm{ext}}=q\left(V_{B}-V_{A}\right)\)
\(40 \mu \mathrm{J}=1 \mu \mathrm{C}\left(V_{B}-V_{A}\right)\)
\(V_{A}-V_{B}=-40 \mathrm{~V}\)

(c) Assertion is correct, Reason is incorrect.