Solve this: Solve this: In the following figure, the e e ric te on Y-axis will be maximum at y equal to Share with your friends Share 52 Sanjay Singh answered this Dear student Electric field at point P is same in magnitude due to both the charges .E=14π∈qr2horizontal component of both the fields cancel out and the vertical component gets added as shown in figureEnet=2Ecosθcosθ=yr and r=y2+d24cosθ=yr=yy2+d24Enet=2Ecosθ=2×14π∈qr2×yrEnet=2×14π∈q×yr3Enet=2×14π∈q×y(y2+d24)32=12π∈×q×y(y2+d24)32for Enet to be maximumdEnetdy=0dEnetdy=q2π∈ddy(y(y2+d24)32)dEnetdy=q2π∈(y2+d24)32×1-y×32×(2y)×(y2+d24)12 (y2+d24)30=(y2+d24)32×1-y×32×(2y)×(y2+d24)12 (y2+d24)3(y2+d24)32×1-y×32×(2y)×(y2+d24)12 =0(y2+d24)12(y2+d24-3y2)=0(y2+d24-3y2)=02y2=d24y=d22so maximum value occur at y=d22Regards 18 View Full Answer