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.