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Electrostatic potential energy

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Question

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.

Moderate
Question

The component of electric field in the x-direction (Ex) is

Solution

Ex=δVδx=3  Vm1;  Ey=δVδy=4  Vm1

ax=qExm=1×106×30.1=3×105  ms2

ay=qEym=1×106×40.1=4×105  ms2

Time taken to cross the x-axis

Using s=ut+12at2  we  get

3.2=12×4×105  t2

or t=400  s

vx=axt=3×105×400=12×103  ms1

vy=ayt=4×105×400=16×103  ms1

v=vx2+vy2=20×103  ms1

Question

The component of electric field in the y-direction (Ey) is

Solution

Ex=δVδx=3  Vm1;  Ey=δVδy=4  Vm1

ax=qExm=1×106×30.1=3×105  ms2

ay=qEym=1×106×40.1=4×105  ms2

Time taken to cross the x-axis

Using s=ut+12at2  we  get

3.2=12×4×105  t2

or t=400  s

vx=axt=3×105×400=12×103  ms1

vy=ayt=4×105×400=16×103  ms1

v=vx2+vy2=20×103  ms1

Question

The time taken to cross the x-axis is

Solution

Ex=δVδx=3  Vm1;  Ey=δVδy=4  Vm1

ax=qExm=1×106×30.1=3×105  ms2

ay=qEym=1×106×40.1=4×105  ms2

Time taken to cross the x-axis

Using s=ut+12at2  we  get

3.2=12×4×105  t2

or t=400  s

vx=axt=3×105×400=12×103  ms1

vy=ayt=4×105×400=16×103  ms1

v=vx2+vy2=20×103  ms1

Question

The velocity of the particle when it crosses the x-axis is 

Solution

Ex=δVδx=3  Vm1;  Ey=δVδy=4  Vm1

ax=qExm=1×106×30.1=3×105  ms2

ay=qEym=1×106×40.1=4×105  ms2

Time taken to cross the x-axis

Using s=ut+12at2  we  get

3.2=12×4×105  t2

or t=400  s

vx=axt=3×105×400=12×103  ms1

vy=ayt=4×105×400=16×103  ms1

v=vx2+vy2=20×103  ms1


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