A simple pendulum with a bob of mass m=1 kg , charge $5\;mu C \;$ and string length l=1m is given a horizontal velocity u in a uniform electric field $\;E=2\times10^{6} V/m\;$ at its bottom most point as shown in figure . It is given that the speed u is such that the particle leaves the circle at point C . Find approx the speed u in m/s (Take g=10 m/s^2)

Question

$(a)\;\sqrt{26} m/s\qquad(b)\;\sqrt{31} m/s\qquad(c)\;6 m/s\qquad(d)\;zero$