11. l If it takes to move a positively charged particle between two charged para
ID: 2289449 • Letter: 1
Question
11. l If it takes to move a positively charged particle between two charged parallel plates, (a) what is the charge on the particle if the plates are connected to a 6.0-V battery? (b) Was it moved from the negative to the positive plate or from the positive to the negative plate?
12. l What are the magnitude and direction of the electric field between the two charged parallel plates in Exercise 11 if the plates are separated by 4.0 mm?
17. IE ll (a) At one third the original distance from a positive point charge, by what factor is the electric potential changed: (1) (2) 3, (3) or (4) 9? Why? (b) How far from a charge is a point with an electric potential value of 10 kV? (c) How much of a change in potential would occur if the point were moved to three times that distance?
Equipotential Surfaces and the Electric Field
30. MC On an equipotential surface (a) the electric potential is constant, (b) the electric field is zero, (c) the electric potential is zero, (d) there must be equal amounts of negative and positive charge.
31. MC Equipotential surfaces (a) are parallel to the electric field, (b) are perpendicular to the electric field, (c) can be at any angle with respect to the electric field.
32. MC An electron is moved from an equipotential surface at to one at It is moving generally in a direction (a) parallel to the electric field, (b) opposite to the electric field, (c) in the same direction as the electric field.
41. l For a point charge, what is the radius of the equipotential surface that is at a potential of 2.50 kV?
42. l A uniform electric field of points vertically upward. How far apart are the equipotential planes that differ by 100 V?
48. l The potential difference involved in a typical lightning discharge may be up to 100 MV (million volts). What is the gain in kinetic energy of an electron accelerated through this potential difference? Give your answer in both electron-volts and joules. (Assume that there are no collisions.)
Explanation / Answer
1)
A) W = q*V
==> q = W/V = 1.6*10^-5/6 = 2.67*10^-6 C
B) from positive plate to negative plate
C) E = V/d = 6/(4*10^-3) = 1500 N/c
2)
F = q*E
m*a = q*E
a = q*E/m = 1.6*10^-19*1000/9.1*10^-31 = 1.758*10^14 m/s^2
v^2-u^2 = 2*a*s
v = sqrt(2*a*s)
= sqrt(2*1.758*10^14*0.1*10^-2)
= 5.93*10^5 m/s
3) I think, q = -1.4*10^-6 C
W = k*q^2/r
= 9*10^9*(1.4*10^-6)^2/8*10^-3
= 2.205 J
Equipotential Surfaces and the Electric Field
30)Equipotential surfaces are surfaces of constant scalar potential.potential is constant
(a) the electric potential is constant
31) Equipotential surfaces (b) are perpendicular to the electric field
32)(b) opposite to the electric field
41 ) kQ / R = 2.5*1000 V
q = 1c
there fore R = 3600000 m
42 ) E*d = 100 d = 100/E
43) we know that energy = qV = 100Mev
energy = 1.2 *10^-11 J
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