Download Feel the Potential of Physics Answers

Feel the Potential of Physics Answers
1. A charge of 5.0 μC is moved from one location to another in an
electric field. 25.0 J of work is done. Find the potential difference.
W
V =
q
25.0 J
=
5.0 x 10-6 C
V = 5.0 x 106 J/C = 5.0 x 106 V
2. When an ion accelerates through a potential difference of 2850 V,
its electric potential decreases by 1.378 x 10-15 J. What is the charge
on the ion?
W
V =
q =
=
q
W
V
1.378 x 10-15 J
2850 V
q = 4.8 x 10-19 C
qV = W
q =
Energy
V
3. If an electron is accelerated through a potential difference of
100 V, what is the final speed of the electron?
W
V =
q
Work = Energy
KE = ½ m v2
W = Energy = q V
= ( 1.6021 x 10-19 C )( 100 V )
= 1.6021 x 10-17 J
3. If an electron is accelerated through a potential difference of
100 V, what is the final speed of the electron?
Energy = KE = 1.6021 x 10-17 J
KE = 1/2 mv2
v =
2 KE
=
m
v = 5.93 x 106 m/s
2 ( 1.6021 x 10-17 J )
9.11 x 10-31 kg
4. A capacitor hooked up to a 6.0-V battery acquires 8.0 μC of charge.
(a) What is the capacitance of the capacitor?
Q = CV
C =
Q
V
C =
=
=
=
=
8.0 x 10-6 C
6.0 V
1.3 x 10-6 C/V
1.3 x 10-6 farads
1.3 x 10-6 F
1.3 μF
4. A capacitor hooked up to a 6.0-V battery acquires 8.0 μC of charge.
(b) If the plates are 0.50 mm apart, what is the electric field
intensity between the plates?
milli-
V
E =
6.0 V
=
d
0.00050 m
E = 12 000 V/m
= 1.2 x 104 V/m
( or 0.50 x 10-3 m )
5. An industrial capacitor can have a capacitance of 2.50 F. Its plates
are 0.75 mm apart. If this capacitor is connected to a
potential difference of 120 V, find
(a) the electric field intensity between the plates
milli-
V
E =
120 V
=
d
0.00075 m
E = 160 000 V/m
= 1.6 x 105 V/m
( or 0.75 x 10-3 m )
5. An industrial capacitor can have a capacitance of 2.50 F. Its plates
are 0.75 mm apart. If this capacitor is connected to a
potential difference of 120 V, find
(b) the charge stored on the capacitor
Q = CV
= ( 2.50 F )( 120 V )
Q = 300 C