Hypothetical Non-Conservative Static Electric Field

Suppose an electric field as depicted in the picture
$$E \propto y$$$$E = ky$$

Now when we take a charge $+q$ along the path $ABCDA$, Then :
$$W_{AB} = qky_2L$$$$W_{BC} = 0 \text{(Since, displacement is perpendicular to the field)}$$$$W_{CD} = -qky_1L$$$$W_{DA} = 0 \text{(Since, displacement is perpendicular to the field)}$$

Total work done :
$$W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$$$$W = qky_2L + 0 - qky_1L + 0$$$$W = qkL (y_2 -y_1) $$

Now if $y_2 \neq y_1$, then $W \neq 0$ Which implies that such particular sort of electric field would act non-conservatively.

Since it can be easily asserted that such a field is impossible as it does not follow that static electric field must be conservative, I have done the derivative for a estabilised field of line charge here.

Non-Conservative Behaviour of Static Electric Field

Inspecting work done over a loop in the field of a line charge by a point charge.







Theory

 Point charge "$+q$" travels in the electric field of line charge "$MN$" having linear charge density "$+\lambda$" in a square loop "$ABCDA$".

$$E_{\bot} = \frac{\lambda}{4\pi\epsilon_0r} [\sin\alpha + \sin\beta]$$$$E_{\parallel} = \frac{\lambda}{4\pi\epsilon_0r} [\cos\alpha - \cos\beta]$$

Work done in $A \rightarrow B$



As movement is done against $E_{\parallel}$ only, we do not consider the presense of $E_{\bot}$.
$$E_{\parallel} = \frac{\lambda}{4\pi\epsilon_0r_1} [\frac{r_1}{\sqrt{(x)^2+(r_1)^2}} - \frac{r_1}{\sqrt{(L-x)^2+(r_1)^2}}]$$
$$W_{AB} = \int_0^{\frac{L}{2}} (qE_{\parallel}(-\hat{i})) \cdot (dx(\hat{i}))$$$$W_{AB} = \int_0^{\frac{L}{2}} -qE_{\parallel}dx$$
$$ W_{AB} = \int_0^{\frac{L}{2}} \frac{-q\lambda}{4\pi\epsilon_0} [\frac{1}{\sqrt{(x)^2+(r_1)^2}} - \frac{1}{\sqrt{(L-x)^2+(r_1)^2}}].dx$$$$W_{AB} = \frac{-q\lambda}{4\pi\epsilon_0} [ ln|\sqrt{(x)^2+(r_1)^2}+x| +ln|\sqrt{(L-x)^2+(r_1)^2}+L-x|]_0^{\frac{L}{2}}$$$$W_{AB} = \frac{-q\lambda}{4\pi\epsilon_0} [ln|(\sqrt{(x)^2+(r_1)^2}+x)(\sqrt{(L-x)^2+(r_1)^2}+L-x)|]_0^{\frac{L}{2}}$$$$W_{AB} = \frac{-q\lambda}{4\pi\epsilon_0} [ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+\frac{L}{2})^2}{(\sqrt{(L)^2+(r_1)^2}+L)(r_1)}|]$$$$W_{AB} = \frac{q\lambda}{4\pi\epsilon_0} [ln|\frac{(\sqrt{(L)^2+(r_1)^2}+L)(r_1)}{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+\frac{L}{2})^2}|]$$


Work done in $B \rightarrow C$



As movement is done against $E_{\bot}$ only, we do not consider the presense of $E_{\parallel}$.

$$E_{\bot} = \frac{\lambda}{4\pi\epsilon_0r} [\frac{\frac{L}{2}}{\sqrt{(\frac{L}{2})^2+(r)^2}} + \frac{\frac{L}{2}}{\sqrt{(\frac{L}{2})^2+(r)^2}}]$$$$E_{\bot} = \frac{\lambda L}{4\pi\epsilon_0} [ \frac{1}{r\sqrt{(\frac{L}{2})^2+(r)^2}}]$$
$$W_{BC} = \int_{r_1}^{r_2} (qE_{\bot}(\hat{j}))\cdot(dr(\hat{j}))$$$$W_{BC} = \int_{r_1}^{r_2} qE_{\bot}dr$$
$$ W_{BC} = \int_{r_1}^{r_2} \frac{q\lambda L}{4\pi\epsilon_0} [\frac{1}{r\sqrt{(\frac{L}{2})^2+(r)^2}}].dr$$$$W_{BC} = \frac{q\lambda\require{cancel}\cancel{L}}{4\pi\epsilon_0}\frac{1}{2\require{cancel}\cancel{L}}[ln|\frac{\sqrt{(\frac{L}{2})^2+(r)^2}-\frac{L}{2}}{\sqrt{(\frac{L}{2})^2+(r)^2}+\frac{L}{2}}|]_{r_1}^{r_2}$$$$W_{BC} = \frac{q\lambda}{8\pi\epsilon_0}[ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_2)^2}-(\frac{L}{2}))(\sqrt{(\frac{L}{2})^2+(r_1)^2}+(\frac{L}{2}))}{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+(\frac{L}{2}))(\sqrt{(\frac{L}{2})^2+(r_1)^2}-(\frac{L}{2}))}|]$$$$W_{BC} = \frac{q\lambda}{8\pi\epsilon_0}[ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+(\frac{L}{2}))^2(r_2)^2}{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+(\frac{L}{2}))^2(r_1)^2}|]$$$$W_{BC} = \frac{q\lambda}{4\pi\epsilon_0}[ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+(\frac{L}{2}))(r_2)}{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+(\frac{L}{2}))(r_1)}|]$$

Work done in $C \rightarrow D$



As movement is done against $E_{\parallel}$ only, we do not consider the presense of $E_{\bot}$.

$$E_{\parallel} = \frac{\lambda}{4\pi\epsilon_0r_2} [\frac{r_2}{\sqrt{(x)^2+(r_2)^2}} - \frac{r_2}{\sqrt{(L-x)^2+(r_2)^2}}]$$
$$W_{CD} = \int_{\frac{L}{2}}^0 (qE_{\parallel}(-\hat{i})) \cdot (dx(\hat{i}))$$$$W_{CD} = \int_{\frac{L}{2}}^0 -qE_{\parallel}dx$$$$W_{CD} = \int_0^{\frac{L}{2}} qE_{\parallel}dx$$
$$ W_{CD} = \int_0^{\frac{L}{2}} \frac{q\lambda}{4\pi\epsilon_0} [\frac{1}{\sqrt{(x)^2+(r_2)^2}} - \frac{1}{\sqrt{(L-x)^2+(r_2)^2}}].dx$$$$W_{CD} = \frac{q\lambda}{4\pi\epsilon_0} [ ln|\sqrt{(x)^2+(r_2)^2}+x| +ln|\sqrt{(L-x)^2+(r_2)^2}+L-x|]_0^{\frac{L}{2}}$$$$W_{CD} = \frac{q\lambda}{4\pi\epsilon_0} [ln|(\sqrt{(x)^2+(r_2)^2}+x)(\sqrt{(L-x)^2+(r_2)^2}+L-x)|]_0^{\frac{L}{2}}$$$$W_{CD} = \frac{q\lambda}{4\pi\epsilon_0} [ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+\frac{L}{2})^2}{(\sqrt{(L)^2+(r_2)^2}+L)(r_2)}|]$$$$W_{CD} = \frac{q\lambda}{4\pi\epsilon_0} [ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+\frac{L}{2})^2}{(\sqrt{(L)^2+(r_2)^2}+L)(r_2)}|]$$


Work done in $D \rightarrow A$



As movement is done against $E_{\bot}$ only, we do not consider the presense of $E_{\parallel}$.

$$E_{\bot} = \frac{\lambda}{4\pi\epsilon_0r} [0+\frac{L}{\sqrt{(L)^2+(r)^2}}]$$$$E_{\bot} = \frac{\lambda L}{4\pi\epsilon_0} [ \frac{1}{r\sqrt{(L)^2+(r)^2}}]$$
$$W_{DA} = \int_{r_2}^{r_1} (qE_{\bot}(\hat{j}))\cdot(dr(\hat{j})$$$$W_{DA} = \int_{r_1}^{r_2} -qE_{\bot}dr$$
$$ W_{DA} = \int_{r_1}^{r_2} \frac{-q\lambda L}{4\pi\epsilon_0} [\frac{1}{r\sqrt{(L)^2+(r)^2}}].dr$$$$W_{DA} = \frac{-q\lambda\require{cancel}\cancel{L}}{4\pi\epsilon_0}\frac{1}{2\require{cancel}\cancel{L}}[ln|\frac{\sqrt{(L)^2+(r)^2}-L}{\sqrt{(\frac{L}{2})^2+(r)^2}+L}|]_{r_1}^{r_2}$$$$W_{DA} = \frac{-q\lambda}{8\pi\epsilon_0}[ln|\frac{(\sqrt{(L)^2+(r_2)^2}-(L))(\sqrt{(L)^2+(r_1)^2}+(L))}{(\sqrt{(L)^2+(r_2)^2}+(L))(\sqrt{(L)^2+(r_1)^2}-(L))}|]$$$$W_{DA} = \frac{-q\lambda}{8\pi\epsilon_0}[ln|\frac{(\sqrt{(L)^2+(r_1)^2}+(L))^2(r_2)^2}{(\sqrt{(L)^2+(r_2)^2}+(L))^2(r_1)^2}|]$$$$W_{DA} = \frac{-q\lambda}{4\pi\epsilon_0}[ln|\frac{(\sqrt{(L)^2+(r_1)^2}+(L))(r_2)}{(\sqrt{(L)^2+(r_2)^2}+(L))(r_1)}|]$$$$W_{DA} = \frac{q\lambda}{4\pi\epsilon_0}[ln|\frac{(\sqrt{(L)^2+(r_2)^2}+(L))(r_1)}{(\sqrt{(L)^2+(r_1)^2}+(L))(r_2)}|]$$

Adding Work

$$W=W_{AB}+W_{BC}+W_{CD}+W_{DA}$$
$$W_{AB} = \frac{q\lambda}{4\pi\epsilon_0} [ln|\frac{(\sqrt{(L)^2+(r_1)^2}+L)(r_1)}{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+\frac{L}{2})^2}|]$$$$W_{BC} = \frac{q\lambda}{4\pi\epsilon_0}[ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+(\frac{L}{2}))(r_2)}{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+(\frac{L}{2}))(r_1)}|]$$$$W_{CD} = \frac{q\lambda}{4\pi\epsilon_0} [ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+\frac{L}{2})^2}{(\sqrt{(L)^2+(r_2)^2}+L)(r_2)}|]$$$$W_{DA} = \frac{q\lambda}{4\pi\epsilon_0}[ln|\frac{(\sqrt{(L)^2+(r_2)^2}+(L))(r_1)}{(\sqrt{(L)^2+(r_1)^2}+(L))(r_2)}|]$$
$$W = \frac{q\lambda}{4\pi\epsilon_0} [ln|\frac{\require{cancel}\cancel{(\sqrt{(L)^2+(r_1)^2}+L)}\require{cancel}\cancel{(r_1)}\require{cancel}\cancel{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+(\frac{L}{2}))}\require{cancel}\cancel{(r_2)}(\sqrt{(\frac{L}{2})^2+(r_2)^2}+\frac{L}{2})^{\require{cancel}\cancel{2}1}\require{cancel}\cancel{(\sqrt{(L)^2+(r_2)^2}+(L))}(r_1)}{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+\frac{L}{2})^{\require{cancel}\cancel{2}}\require{cancel}\cancel{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+(\frac{L}{2}))}\require{cancel}\cancel{(r_1)}\require{cancel}\cancel{(\sqrt{(L)^2+(r_2)^2}+L)}\require{cancel}\cancel{(r_2)}\require{cancel}\cancel{(\sqrt{(L)^2+(r_1)^2}+(L))}(r_2)}|]$$$$W = \frac{q\lambda}{4\pi\epsilon_0} [ln|\frac{(\sqrt{(\frac{L}{2})^2+(r_2)^2}+\frac{L}{2})(r_1)}{(\sqrt{(\frac{L}{2})^2+(r_1)^2}+\frac{L}{2})(r_2)}|]$$$$W = \frac{q\lambda}{4\pi\epsilon_0} [ln|\frac{(\sqrt{(\frac{L}{2r_2})^2+1}+\frac{L}{2r_2})}{(\sqrt{(\frac{L}{2r_1})^2+1}+\frac{L}{2r_1})}|]$$

Conclusion

So if $r_1 \neq r_2$, then $W \neq 0$ which indicates that even static electric field behaves non-conservatively in some cases.

This phenomenon is extremely easy to show in hypothetical electric fields as I have shown here.

Improving transformers ?

The traditional tranformers suffer a lot of losses in various forms, some of them are as follows :
1. Flux leakage
2. Eddy currents
3. Hysteresis

But as we are taught induction we come across the attached figure of two inductors placed coaxially various times. My point is that if we would actually place inductors in this way, we would have the following benefits :

Note : taking inner coil as primary

1. NO NEED OF CORE : As the flux of inner coil passes from outer on its own, no core is needed for transferring flux from primary to secondary
2. NO FLUX LEAKAGE : As entire flux of inner inductor passes from outer, no flux is leaked.
3. NO EDDY CURRENTS: As no core is used no eddy currents are generated.
4. NO HYSTERESIS : Similarly due to absense of core, no loss due to hysteresis is encountered.

So what really is the problem in creating transformers in such a way to reduce losses much more than the current status ?

All relevant opinions and comments appreciated