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Q.

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 is

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a

−3 Vm−1

b

4 Vm−1

c

5 Vm−1

d

8 Vm−1

e

3 Vm−1

f

-4 Vm−1

g

5 Vm−1

h

8 Vm−1

i

20 s

j

40 s

k

200 s

l

400 s

answer is , , .

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Detailed Solution

Ex=−δVδx=−3Vm−1; Ey=−δVδy=−4Vm−1ax=qExm=−1×10−6×30.1=−3×10−5ms−2ay=qEym=−1×10−6×40.1=−4×10−5ms−2Time taken to cross the x-axisUsing s=ut+12at2 we get 3.2=12×4×10−5×t2or t=400svx=axt=−3×10−5×400=12×10−3ms−1vy=ayt=−4×10−5×400=16×10−3ms−1v=vx2+vy2=20×10−3ms−1Ex=−δVδx=−3Vm−1; Ey=−δVδy=−4Vm−1ax=qExm=−1×10−6×30.1=−3×10−5ms−2ay=qEym=−1×10−6×40.1=−4×10−5ms−2Time taken to cross the x-axisUsing s=ut+12at2 we get 3.2=12×4×10−5×t2or t=400svx=axt=−3×10−5×400=12×10−3ms−1vy=ayt=−4×10−5×400=16×10−3ms−1v=vx2+vy2=20×10−3ms−1Ex=−δVδx=−3Vm−1; Ey=−δVδy=−4Vm−1ax=qExm=−1×10−6×30.1=−3×10−5ms−2ay=qEym=−1×10−6×40.1=−4×10−5ms−2Time taken to cross the x-axisUsing s=ut+12at2 we get 3.2=12×4×10−5×t2or t=400svx=axt=−3×10−5×400=-12×10−3ms−1vy=ayt=−4×10−5×400=-16×10−3ms−1v=vx2+vy2=20×10−3ms−1
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