Potential at the centre,
\(V=\frac{1}{4 \pi \Sigma 0}\left(\frac{q}{r}\right)\)
\(=9 \times 10^{9} \times 10^{-8} \times \frac{60}{360} \times 2 \pi r\)
\(=9 \times 10^{+9} \times 10^{-8} \times \frac{2 \times 3.14 \times 2}{6}=188.4 \mathrm{~V}\) 

(i) Electric field between the plates is given by
\(E=\frac{\Delta V}{\Delta x}=-\frac{\left(V_{B}-V_{A}\right)}{1 \times 10^{-2}}\)
\(=\frac{-(0-10)}{10^{-2}}=10^{3} \mathrm{~V} / \mathrm{m}\)
It is directed from A to B
(ii) Work done in moving a charge from A to B,
W_{A\(\rightarrow\)B} = q(\(\triangle\) V) = 20 x 10^-6 (V_B - V_A)
= 20 x 10^-6 (0 -10)
= - 20 x 10^-5 J

(i) The direction of electric field is perpendicular to their equipotential surface. So,  the direction of electric field is along X-axis as its length should be perpendicular to equipotential surface lying in YZ- plane.
(ii) Length of the dipole - 2b
As dipole's axis is along the Y-axis.
\(\therefore \)  Electric dipole moment,
p = q (2b)\(\overset { \wedge }{ j } \)
Electric field, E = E\(\overset { \wedge }{ i } \)
\(\because \)   \(\tau \) = p x E
= q(2b)\(\overset { \wedge }{ j } \) x E\(\overset { \wedge }{ i } \)
= +2 qbE \((\overset { \wedge }{ j } \times\overset { \wedge }{ i } )\)
= 2 qbE (-\(\overset { \wedge }{ k } \))
\(\therefore \) Torque, |\(\tau \)| = 2 qbE.

\(Here,q=25\times { 10 }^{ -9 }C,\ 2a=6m,r=4m\)
\(p=q(2a)=25\times { 10 }^{ -9 }\times 6=1.5\times { 10 }^{ -7 }C-m\)
\(Now,\ { E }_{ axial }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { 2pr }{ { ({ r }^{ 2 }-{ a }^{ 2 } })^{ 2 } } \)
\(=\frac { 9\times { 10 }^{ 9 }\times 2\times 1.5\times { 10 }^{ -7 }\times 4 }{ { ({ 4 }^{ 2 }-{ 3 }^{ 2 }) }^{ 2 } } =\frac { 2700\times 4 }{ 49 } \)
\({ E }_{ axial }=220.4{ NC }^{ -1 }\)
\({ E }_{ equatorial }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { p }{ { ({ r }^{ 2 }+{ a }^{ 2 } })^{ 3/2 } } \)
\(=\frac { 9\times { 10 }^{ 9 }\times 1.5\times { 10 }^{ -7 } }{ { ({ 4 }^{ 2 }+{ 3 }^{ 2 }) }^{ 3/2 } } =\frac { 1350 }{ 125 } \)
\(=10.8{ NC }^{ -1 }\ \)

According to the formula, E.dr = dV, the value of dV = 0 at each point of equipotential surface.
\(\therefore \) E.dr = 0
so, the angle between electric field vector and distance vector will be 90^o . Thus, the electric field is always normal to the plane passing through AB.