A metallic sphere of radius 2.0 cm is charged with +5.0-μC+5.0-μC charge, which spreads on the surface of the sphere uniformly.
1 answer:
Answer:
Explanation:
Potential due to a charged metallic sphere having charge Q and radius r on its surface will be
v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is
v = k Q / R
a ) On the surface of the shell , potential due to positive charge is
V₁ =
On the surface of the shell , potential due to negative charge is
V₁ =
Total potential will be zero . they will cancel each other.
b ) On the surface of the sphere potential
=
= 22.5 x 10⁵ V
On the surface of the sphere potential due to outer shell
=
= -9 x 10⁵
Total potential
=( 22.5 - 9 ) x 10⁵
= 13.5 x 10⁵ V
c ) In the space between the two , potential will depend upon the distance of the point from the common centre .
d ) Inside the sphere , potential will be same as that on the surface that is
13.5 x 10⁵ V.
e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.
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Answer:
A) i) E =α/ [2πrL(εo)]
ii) E=0
iii) E = α/(πrεo)
The graph between E and r for the 3 cases is attached to this answer ;
B) i) charge on the inner surface per unit length = - α
ii) charge per unit length on the outer surface = 2α
Explanation:
A) i) For r < a, the charge is in the cavity and takes a shape of the cylinder. Thus, applying gauss law;
EA = Q(cavity) / εo
Now, Qcavity = αL
So, E(2πrL) = αL/εo
Making E the subject of the formula, we have;
E =α/ [2πrL(εo)]
ii) For a < r < b; since the distance will be in the bulk of the conductor, therefore, inside the conductor, the electric field will be zero.
So, E=0
iii) For r > b; the total enclosed charge in the system is the difference between the net charge and the charge in the inner surface of the cylinder.
Thus, Qencl = Qnet - Qinner
Qinner will be the negative of Qnet because it should be in the opposite charge of the cavity in order for the electric field to be zero. Thus;
Qencl = αL - (-αL) = 2αL
Thus, applying gauss law;
EA = Qencl / εo
Thus, E = Qencl / Aεo
E = 2αL/Aεo
Since A = 2πrL,
E = 2αL/2πrLεo = α/(πrεo)
B) i) The charge on the cavity wall must be the opposite of the point charge. Therefore, the charge per unit length in the inner surface of the tube will be - α
ii)Net charge per length for tube is +α and there is a charge of - α on the inner surface. Thus charge per unit length on the outer surface will be = +α - (- α) = 2α
Answer:
Here's what I get
Explanation:
A. Distance between A and B.
h = -½gt²
The stones go faster the farther they fall.
Stone A has already reached 5 m when B is released.
When B reaches 5 m, A has dropped further and is falling even faster.
The distance between the stones increases with time.
Figure 1 shows this effect in a graph of height vs. time.
B. Speed of Stone B
v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²
v = 9.9 m/s
The stone is travelling at 9.9 m/s when it reaches 5 m.
C. Velocity vs time
v = -gt
Both stones accelerate at the same rate.
When Stone B has reached 10 m at time t, Stone A is falling much faster.
Fig. 2 shows this in a graph of velocity vs time.
