Question

A uniform electric field of magnitude 375N/C pointing in the positive x-direction acts on an electron, which is initially at rest.After the electron has moved 3.20cm,what is

(a)the work done by the field on the electron
(b)the charge in potential energy associated with the electron
(c)the velocity of the electron (mass of electron=9.11^-31 kg)

Answer 1

a) W = - 1.92 10⁻¹⁸ J, b) U = 1.92 10⁻¹⁸ J, c) v = 2.05 10⁶ m / s

Step-by-step explanation:

a) Work is defined by

W = F. x

the electric force is

F = q E

we substitute

W = q E x

where the displacement is parallel to the electric field

all quantities must be in the SI system

x = 3.20 cm = 0.0320 m

W = - 1.6 10-19 375 0.0320

W = - 1.92 10⁻¹⁸ J

b) potential energy

U =
`k (q_1q_2)/(r)`

the electrical power is

V = k q₁ / r

we can write this potential as a function of the electric field

V = -E x

we substitute

U = -q E x

U = - (-1.6 10⁻¹⁹) 375 0.0320

U = 1.92 10⁻¹⁸ J

Observe that the variation of the red potential is equal to the electrical work

c) let's use conservation of energy

starting point

Em₀ = U = e E x

final point

Em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

e E x = ½ m v²

v² = 2 e E x / m

v =
`\sqrt{ \frac{{2 \ 1.6 10^(-19) \ 375 \ 0.0320 } }{9.1 \ 10^(-31) } }`

v =
`\sqrt{4.2198 \ 10^(12)}`

v = 2.05 10⁶ m / s