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Dear Student,

$$\begin{gathered} \left(a\right) \\ general expression of E. Potential due to a dipole \\ V=\frac{1}{4\pi {\epsilon }_0}\frac{p\cos \theta }{r^2} \\ at axial point \theta =0 \\ or, V_a=\frac{p}{4\pi {\epsilon }_0{r}^2} \end{gathered}$$

(b)  Electrical force is directed from A to B  thus work done= FdS= W (say)
while moving from B to A dS= -tive now or work done=-W
​since the field is uniform, and it is conservative we have
ratio of work done A-B and work done B-A= -1

(c)
 $$\begin{gathered} We have electric field E=-\frac{∆V}{∆x} ie- gradient of potential \\ in other words, E=-\frac{Potential}{difference between two lines} or density of Potential \\ Thus for Electric Field to be greater while Potential is same, we need to have \\ seperation between them very small. \end{gathered}$$

(d) The following graph can be visualised as following

$$\begin{gathered} V=slope\times x where x=\frac{1}{r} \\ for a +tive charge we have V=\frac{kQ}{r}\Rightarrow \left(KQ\right)\frac{1}{r}\Rightarrow +tive slope \times x \\ for -tive charge V=-tive slope \times x \\ also slope \propto Q ( magnitude of charge ) as K=cons\tan t \\ from the graph we can see, \\ {Q}_1 has positive slope thus its +tively charged \\ {Q}_{2 }is -tively charged \\ also slope of Q_2>Q_1 thus magnitude of Q_2 is greater \end{gathered}$$

Regards.