GO A positron (charge +e , mass equal to the electron mass) is moving at 1.0 × 10 7 m/s in the positive direction of an x axis when, at x = 0, it encounters an electric field directed along the x axis. The electric potential V associated with the field is given in Fig. 24-57. The scale of the vertical axis is set by V s = 500.0 V. (a) Does the positron emerge from the field at x = 0 (which means its motion is reversed) or at x = 0.50 m (which means its motion is not reversed)? (b) What is its speed when it emerges? Figure 24-57 Problem 54.
GO A positron (charge +e , mass equal to the electron mass) is moving at 1.0 × 10 7 m/s in the positive direction of an x axis when, at x = 0, it encounters an electric field directed along the x axis. The electric potential V associated with the field is given in Fig. 24-57. The scale of the vertical axis is set by V s = 500.0 V. (a) Does the positron emerge from the field at x = 0 (which means its motion is reversed) or at x = 0.50 m (which means its motion is not reversed)? (b) What is its speed when it emerges? Figure 24-57 Problem 54.
GO A positron (charge +e, mass equal to the electron mass) is moving at 1.0 × 107 m/s in the positive direction of an x axis when, at x = 0, it encounters an electric field directed along the x axis. The electric potential V associated with the field is given in Fig. 24-57. The scale of the vertical axis is set by Vs = 500.0 V. (a) Does the positron emerge from the field at x = 0 (which means its motion is reversed) or at x = 0.50 m (which means its motion is not reversed)? (b) What is its speed when it emerges?
-19
Suppose an electron (q = -e = - 1.6 × 10 C,m=9.1×10¬
-31
kg) is accelerated from rest through a potential difference of Vab
= +5000
V. Solve for the final speed of the electron. Express numerical answer in two significant figures.
The potential energy U is related to the electron charge (-e) and potential Vab
related by the equation:
U =
Assuming all potential energy U is converted to kinetic energy K,
K + U = 0
K= -U
1
Since K= mv² and using the formula for potential energy above, we arrive at an equation for speed:
v = ( 2eV/m
1/2
Plugging in values, the value of the electron's speed is:
v= 4.1
x 107 m/s
59 In Fig. 24-60, a charged particle
(either an electron or a proton) is moving
rightward between two parallel charged
plates separated by distance d = 2.00 mm.
The plate potentials are Vi = -70.0 V and
V2 = -50.0 V. The particle is slowing from
an initial speed of 90.0 km/s at the left i
plate. (a) Is the particle an electron or a
proton? (b) What is its speed just as it
reaches plate 2?
%3D
V2
Figure 24-60
Problem 59.
A mass has a charge of q =
3e, where e is the charge on an electron.
(a) Determine the electric potential (in V) due to the charge at a distance r = 0.220 cm from the charge.
V
(b) Determine the electric potential difference (in V) between a point that is 3r away and this point, that is V(3r) – V(r).
V
r
away and this point, that is V
3
r
(c) Determine the electric potential difference (in V) between a point that is
) - ver).
3
V
(d) How would the answers change if the electrons are replaced by protons?
The sign of answer (a) would change.
The sign of answer (b) would change.
O The sign of answer (c) would change.
O The sign of all answers would change.
O All answers would remain the same.
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