Q20._ElectricChargeForceAndField

Transcription

Q20._ElectricChargeForceAndField
Q20. Electric Charge Force and Field
1. A small object has charge Q.
Charge q is removed from it
and placed on a second small object.
placed 1 m apart.
The two objects are
For the force that each object exerts on
the other to be a maximum, q should be :
1.
2Q
2.
Q
3.
Q/2
4.
Q/4
5.
0
Qq
q
q Q  q 
f k
r2
df
k
 2 Q  2q   0
dq r

Q
q
2
d2 f
k
 2 2  0
2
dq
r
( max )
2. Two identical conducting spheres A and B carry equal charge.
They are separated by a distance much larger than their
diameters.
A third identical conducting sphere C is uncharged.
Sphere C is first touched to A, then to B, and finally removed.
As a result, the electrostatic force between A and B, which was
originally F, becomes :
1.
F/2
2.
F/4
3.
3F / 8
4.
F / 16
5.
0
A, Q
B, Q
C, 0
Spheres identical  Charges are shared equally after contact
After C touching A :
After C touching B :
Q
Q A  QC 
2
1Q
 3
QB  QC    Q   Q
2 2
 4
k Q 3Q 3
 F
Final force between A & B : F   2
r 2 4 8
3. Particles 1, with charge q1, and 2, with a charge q2, are on
the x axis, with particle 1 at x = a and particle 2 at x  –2a.
For the net force on a third charged particle at the origin to
be zero, q1 and q2 must be related by q2  :
1.
2 q1
2.
4 q1
3.
– 2 q1
4.
– 4 q1
5.
– q1 / 4
2a
q
q2
a
q1
 q
q2 
1
k q3   2 
0
2 
 a


2
a





q2  4 q1
4. A particle with charge 2 C is placed at the origin. An
identical particle, with the same charge, is placed 2 m from
the origin on the x axis, and a third identical particle, with the
same charge, is placed 2 m from the origin on the y axis. The
magnitude of the force on the particle at the origin is
1.
9.0  10–3 N
2.
6.4  10–3 N
3.
1.3  10–2 N
4.
1.8  10–2 N
5.
3.6  10–2 N
y=2m
q = 2C
x=0
q = 2C
x=2m
q = 2C
9  10  2  10
9
2
 2
2

6 2
2
 1.27  10 N
5. A charge Q is spread uniformly along the circumference of a
circle of radius R.
A point particle with charge q is placed at
the center of this circle.
The total force exerted on the
particle q can be calculated by Coulomb's law :
1.
just use R for the distance
2.
just use 2R for the distance
3.
just use 2p R for the distance
4.
result of the calculation is zero
5.
none of the above
Q
q
R
f  0 by symmetry.
6. A particle with charge Q is on the y axis a distance a from the
origin and a particle with charge q is on the x axis a distance d
from the origin.
The value of d for which the x component
of the force on the second particle is the greatest is :
1.
0
2.
a
3.
a2
4.
a/2
5.
a/2
Q
r
a
1 d
d
f k qQ 2 k qQ
3/2
2
2
r r
a  d 
f
d
q


2
d
d
df
1
3  

0
k qQ

3/2
5/2
dd
2 a2  d 2  
 a2  d 2 



a
d
2
7. A particle with a charge of 5  10–6 C and a mass of 20 g
moves uniformly with a speed of 7 m/s in a circular orbit
around a stationary particle with a charge of 5  10–6 C.
The radius of the orbit is :
1.
0
2.
0.23 m
3.
0.62 m
4.
1.6 m
5.
Orbit is impossible
Q = 5  10–6 C
m = 20 g
v = 7 m/s
q = 5  10–6 C
R
Circular motion requires a centripetal force.
Coulomb force between the particles is
however repulsive, and hence centrifugal.