A charged ball of mass 9 kg is suspended from a string in a uniform electric field $\overrightarrow{\mathrm{E}}=(3\hat{\mathrm{i}}+5\hat{\mathrm{j}})\times {10}^5\mathrm{N}/\mathrm{C}$. The ball is in equilibrium with $\mathrm{\theta }={37}^{\circ }$. If direction of electric field is reversed, find the new equilibrium position of the ball. Give your answer in terms of angle made by string with vertical. Take $\mathrm{g}=10{\mathrm{ms}}^{-2}$.