The electric potential varies in space according to the relation V=3x+4y. A particle of mass 0.1 kg starts from rest from point (2,3.2) under the influence of this field. The charge on the particle is +1  μC. Assume V and (x, y) are in SI units.The component of electric field in the x-direction (Ex) isThe component of electric field in the y-direction (Ey) isThe time taken to cross the x-axis isThe velocity of the particle when it crosses the x-axis is

Ex=−δVδx=−3  Vm−1;  Ey=−δVδy=−4  Vm−1ax=qExm=−1×10−6×30.1=−3×10−5  ms−2ay=qEym=−1×10−6×40.1=−4×10−5  ms−2Time taken to cross the x-axisUsing s=ut+12at2  we  get3.2=12×4×10−5  t2or t=400  svx=axt=−3×10−5×400=12×10−3  ms−1vy=ayt=−4×10−5×400=16×10−3  ms−1v=vx2+vy2=20×10−3  ms−1Ex=−δVδx=−3  Vm−1;  Ey=−δVδy=−4  Vm−1ax=qExm=−1×10−6×30.1=−3×10−5  ms−2ay=qEym=−1×10−6×40.1=−4×10−5  ms−2Time taken to cross the x-axisUsing s=ut+12at2  we  get3.2=12×4×10−5  t2or t=400  svx=axt=−3×10−5×400=12×10−3  ms−1vy=ayt=−4×10−5×400=16×10−3  ms−1v=vx2+vy2=20×10−3  ms−1Ex=−δVδx=−3  Vm−1;  Ey=−δVδy=−4  Vm−1ax=qExm=−1×10−6×30.1=−3×10−5  ms−2ay=qEym=−1×10−6×40.1=−4×10−5  ms−2Time taken to cross the x-axisUsing s=ut+12at2  we  get3.2=12×4×10−5  t2or t=400  svx=axt=−3×10−5×400=12×10−3  ms−1vy=ayt=−4×10−5×400=16×10−3  ms−1v=vx2+vy2=20×10−3  ms−1Ex=−δVδx=−3  Vm−1;  Ey=−δVδy=−4  Vm−1ax=qExm=−1×10−6×30.1=−3×10−5  ms−2ay=qEym=−1×10−6×40.1=−4×10−5  ms−2Time taken to cross the x-axisUsing s=ut+12at2  we  get3.2=12×4×10−5  t2or t=400  svx=axt=−3×10−5×400=12×10−3  ms−1vy=ayt=−4×10−5×400=16×10−3  ms−1v=vx2+vy2=20×10−3  ms−1