An arrangement of source charges produces the electric potential v=5000x2v=5000x2 along the x-axis, where v is in volts and x is in meters. what is the maximum speed of a 1.0 g, 10 nc charged particle that moves in this potential with turning points at ± 8.0 cm?

The maximum speed of the particle in arrangement of source charges produces 32 volts electric potential is 0.025 meter per second.

Electric potential energy is the energy which is required to move a unit charge from a point to another point in the electric field.

It can be given as,

Here, (q) is the charge and (V) is the electric potential difference.

Given infroamtion-

The electric potential producers by the arrangement of source charges is given by,

The mass of the particle is 1.0 gram.

The charge of the particles 10 nC.

As, the electric potential producers by the arrangement of source charges is given by,

At the x equal to 8 cm or 0.08 m, the equation become,

Thus the potential difference at the is 32 volts.

The electric potential energy of the particle is,

Now the electric potential energy is equal to the kinetic energy of the particle. Thus,

Thus the maximum speed of particle is 0.025 meter per second.

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v = 0.025 m/s

Explanation:

Given that the voltage is

now at x = 0

also we have at x = 8 cm

now change in potential energy of the charge is given as

now by mechanical energy conservation law

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s