V = V1 + V2
Q = C1V = 6 x 10^-6  x 2 = 12\(\mu \)C
As C₂ is in series same amount of charge will also flow through it now V₂ = Q/C₂ = (12 x 10^-6) /(12 x 10^-6) = 1 volt
Total Battery voltage, V = 2 + 1 = 3 Volt

q \(\rightarrow\)_______Q
1/2 mv2 = kQq/r
Or, v2 a1/r
Or, r a 1/v²
Or, r' = r/4

10^-4Nm

Given: \(q_{1}=4 \mu \mathrm{C}=4 \times 10^{-6} \mathrm{C}\)
\(q_{2}=-6 \mu \mathrm{C}=-6 \times 10^{-6} \mathrm{C}, r=20 \mathrm{~cm}\)
Let the electric potential be zero at a point P, a distance x (in cm) from q₁ Then
\(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{x}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{(r-x)}=0\)
\( \therefore \frac{4}{x} =-\frac{(-6)}{(20-x)} \)
\(\Rightarrow 4(20-x=6 x \)
\(x =8 \mathrm{~cm}\)
i.e. 8 cm from 4 μC charge.

(i) \({{10}^{-9}}C\)
(ii) r = 0.5 m

q = 0.48C

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

Here q = 100C
Potential difference between cloud and the earth V = 10⁷V
Energy dissipated W = qV = 100 \(\times 10^7=10^9J\)

\(Here, q=\pm20\mu C=\pm20\times10^{-6}C\)
\(2a=1 cm=10^{-2}m, r=50cm={1\over 2}m\)
As 2a<<r, therefore, intensity on equatorial line of short dipole is
\(E={1\over 4\pi\epsilon_o}{P\over r^3}={q\times 2a\over 4\pi\epsilon_or^3}\)
\(={9\times 10^9\times 20\times10^{-6}\times10^{-2}\over(1/2)^3}\)
\(E=1.44\times 10^4N/C\)

Here, \(q=1\mu C=10^{-6}C\)
2a = 1cm = 10^-2m , r = 10cm = 10^-1m
On the axial line of dipole,
\(E={2|\overrightarrow { p } |r\over 4\pi\epsilon_o(r^2-a^2)^2}={2q\times 2a\times r\over 4\pi\epsilon_o(r^2-a^2)^2}\)
\(={9\times10^9\times 2\times 10^{-6}\times 10^{-2}\times 10^{-1}\over(10^{-2}-0.25\times 10^{-4})^2}\)
\(={18\over 10^{-4}(1-0.0025)^2}\)
\(E={18\times 10^4\over0.9975\times 0.9975}=18\times10^4N/C\)

Here, m = 10^-3kg, \(q=1.6\times 10^{-19}C\)
Force on water drop due to electric field = Weight of water drop
qE = mg
\(E={mg\over q}={10^{-3}\times9.8\over1.6\times 10^{-19}}=6.125\times 10^{16}N/C\)