Question

A uniform electric field of magnitude E=270V/m is directed in the positive x direction. A proton moves from the origin to the point (x, y)=(20.0cm, 50.0cm).

a) Through what potential difference does the charge move?

b) What is the change in the potential energy of the charge field system?

c) An electron is released at rest at the origin and it moves in the +x direction. What would be its speed, V_f, after the electron is released from rest and travels distrance d in the uniform E field? Please derive the express of v_f as a function of the given quantities: E, d, e (the elementary charge) and m (the mass of the electron).

Answer 1

Solution:

Part (a) A uniform electric field E_x = 270V/m is in the positive X direction.

A proton moves from the origin (0.0cm, 0.0cm) to a point p (20.0cm, 50.0cm).

The net displacement in positive X direction is x = 20 cm = 0.2 m.

The electric potential difference between the origin and the given point is obtained by

ΔV = V_f – V_i = -E^.x

ΔV = - 270V/m.0.2m

ΔV = - 54.0 V

The proton move through the potential difference of -54 Volts

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Part (b)

The change in the potential energy of the charge-filed system is given by,

ΔU = ΔV*q

ΔU = (-54.0V)*(1.6*10^-19C)

ΔU = -8.64*10^-18 J

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Part (c)                   Mass of electron m = 9.31*10^-31kg

An electron at rest is released in an uniform electric field E.

The initial velocity v_i = 0 m/s. Thus initial kinetic energy of the electron is K_i = (1/2)m*v_i²

K_i = (1/2)m*(0m/s)²

K_i = 0 J

Let the initial potential energy is U_i = 0J (as the we can choose arbitrary zero potential )

Thus total intial energy is Ti = K_i + U_i

T_i = 0 J

When the electron moves a distance d in uniform it looses it potential energy and gains the equal amount of kinetic energy.

Let after travelling the distance of d m in an electric field E, the final potential energy of the electron be,

U_f = ΔV*q

U_f =(-E.d)*(e)

The final kinetic energy of electron K_f = (1/2) m*v_f²

The total final energy T_f = U_f + K_f

But initial and final total energy are same

T_f = T_i

U_f + K_f = U_i + K_i

(-E.d)*e + (1/2) m*v_f² = 0

v_f² = 2*E*d*e/m

taking square root of above equation we have,

v_f = √(2Ede/m)