II. LA VALORACIÓN DE LOS FACTORES PLUS COMO INDICIOS
1. Los indicios en la legislación civil colombiana
ds is a small instantaneous displacement along the path.
Fe and Fext are the electrostatic and external forces respectively.
Fext is almost equal and opposite to Fe since we are to move the charge q slowly (quasi statically) so our applied force should be such that it appears to be at almost equilibrium at all instant but over a prolonged period, the charge would appear to have been shifted.
dr = ds cosq is the displacement along the radial vector, i.e., displacement in the direction of force.
\ Work done by external agent, Wext=
∫
dwext= −∫
( Fext)( )dr= −
∫
F dre = −∫
kqQr dr k=r r
2
1 0
2 1
; where 4πε = − −
kqQ
r r 1 r
1
2 \ W kqQ r
kqQ
ext= − r
2 1
This work clearly is only initial and final positions dependent but independent of path. Hence we conclude that the external agent was performing work against the conservative electrostatic forces.
Therefore we can associate PE with it, which by definition is,
Wext = DPE
\ U2 – U1 = kqQ r
kqQ
2 r1
−
\ PE at a separation r becomes, U(r) = kqQ
r +U∞ where U∞ indicates the potential energy at infinite separation, that is conventionally taken to be zero until and unless specified.
So, if U∞= 0 then U r kqQ
( ) = r
\ U∝ 1, so can we conclude that by increasing r
C NCEPT C RE
Electrostatic Potential Energy and
Grounding of Conductors
separation, PE decreases? No, not if the charges are unlike. If qQ < 0, then U kqQ= r < 0, so increasing r, gives a less negative PE, so obviously PE increases.
Forget this trap of –ve and +ve. Let me give you a Golden rule!
Whenever you release any system and allow them to move spontaneously, it always moves towards lower PE configuration.
So, if q and Q are both +ve or both –ve, due to repulsion they would spontaneously move away which mean if r increases, PE decreases.
If q and Q are unlike, they attract and move closer so if r decrease, PE decreases, hence if r increases, PE increases.
Similarly, if we use the definition of PE as work done by external agent to build the system assuming they were initially placed far away.
r1 = ∞ (far away) PE of a System of three Point charges
Assume, that the charges are far away from each other.
When we bring q1 from infinity to its position, no work would be done since there are no charges in vicinity which would exert force on the
r12 r31 r23 q3 q2
q1
charge. Now keeping q1 fixed q2 is brought till separation
r12, so work would be required to be done against the force of q1. Now keeping both q1 and q2 fixed, q3 is moved for which work would again be done against forces of q1 and q2.
A close introspection shows us that we need to form as many two charge pairs that we can and then for each two charge pair we write kq q
r
i j ij
.
For example consider four identical charges placed at the vertices of a square. We have 4 edge pairs and 2 diagonal pairs.
\ Usystem = 4 2
Let us see few examples of it before moving to next section.
E1. Consider two point charges of given charges q1 and q2 and masses m1 and m2 which are released from an initial separation r0. Find their relative velocity when the separation between them becomes large.
Sol.: q1 r0 q2
q1 q2
v1 v2
Notice that on combined mass system, (there are no external forces, so linear momentum has to be conserved which means, m1v1 = m2v2 = p (say)
\ Using energy conservation (DE = 0), KEi + PEi = KEf + PEf
Alternatively, in COM (centre of mass) frame KEsystem = 1 shown in the figure.
Find the work required to be done to do so?
Sol.: We observe that the seperation of Q from q1 does not change, so no work would be required to be done for it, so consider it to be not existing for us.But for q2, ri = 5a and rf = 7a
Now, let us define potential difference between two points. Potential difference between two points A and B is defined as minimum amount of work required to be done per unit test charge in moving the test charge from point A to point B.
From work energy theorem (WET), Welectric field + Wext agent = DKE = 0
Therefore the line integral of E
over any path, taken with a negative sign gives us the potential difference between the two points.
Few note worthy points
Why would anyone wish to calculate potential
z
difference between two points?
Since if VB – VA is known, multiplying it with a charge q gives the work required to be done in moving the charge form A to B, and hence it also gives the change in potential energy.
What does negative in the expression of
z VB – VA on
RHS indicate?
Suppose we choose the points A and B in the direction of a field as below.
\ Electric field points in the direction of decreasing potential. coordinates of the respective points A and B.
Specifically, as a special case, if electric field is uniform, that is, constant in magnitude as well as direction, Ex, Ey and Ez will be constant in which case they can be taken out of integration symbol, as below.
V VB A E dx E dy E dzx
Instead of memorizing the formula, let us try to decode the expression.
Ex Dx indicates component of E
along x axis multiplied with distance between the two points along x axis and this gives us –ve of the potential difference between the two points. Let us find a logical way to do this.
Electric field along any direction multiplied with the separation between any two points in this direction gives the magnitude of potential difference between the two points. Whether it is positive or negative can be decided by the direction of electric field.
|VA – VB| = (E cosq) l
What do we mean by potential at a point?
z
The definition says that it refers to the minimum work done by external agent per unit test charge in bringing the test charge from infinity to the point
For example, potential due to a point charge Q at a distance r.
At P, P
Where V(∞) indicates potential due to +Q at infinity.
This value is generally taken to be zero until and unless specified.
One thing, we should remember here is that the potential at any point is dependent upon the choice of zero potential but potential difference between any two points is independent of this choice.
So if V(∞) = 0, then
V kQ
P = r , which is the potential for a point charge Q at a point P at a distance r.
Unlike electric field which is a vector quantity, electric potential is a scalar quantity, hence the positioning of the charge is immaterial as long as the separation remains unchanged.
due to multiple point charges is a scalar sum of potential due to individual point
The locus of all such point in space which are at identical potential is said to be equipotatial surface.
There cannot be any non-zero component of electric field along equipotential surface else it would create a potential difference which is a contradiction.
Hence E
will always be perpendicular to equipotential surfaces.
Saying a surface is equipotential is different from saying that the entire region of a space is equipotential. In the 1st case, E
may or may not be present but if it is present, it has to be perpendicular to the surface but in 2nd case E
= 0 everywhere in the region else in whichever direction E
≠ 0 along this direction potential difference would be created which is a contradiction.
This also means that the entire conductor under electrostatic condition is equipotential. Since if E
inside the bulk, they will drift free electrons inside ≠ 0 the bulk and if E
is not perpendicular to surface, again drifting of free electrons along the surface would happen, both of which are contradictions of electrostatic condition.
Some standard results
Potential due to a charged ring at a point on its axis Consider the ring to be Q
x P
2 2
R x
R made up of large number
of elemental point charges dq. Each dq element is at
Potential due to a uniformly charged thin spherical shell We know from shell theorem that the field of a uniformly charged sphere matches with a point charge for external points so the results of potential will match too, i.e., for external points, we can assume the entire charge on sphere to be concentrated at the centre.
But for internal points E
= 0 , hence potential will become constant and equal to that of the surface since no further work is required to be done in moving the test charge from surface to internal point.
+ +
Potential due to a uniformly charged solid sphere For external points we can again imagine the entire charge to be concentrated at the centre.
But for internal points let us see how
P is an internal point at a distance r from the centre (r < R) we have divided the entire sphere into two parts namely 1 and 2 . large number of thin concentric shells.
Charge dq in the elemental shell of radius x and Potential at centre, for r = 0,
V kQ
C= 3 R 2
and potential at surface, V kQ
S = R
\ VC=3VS
2 , this result is applicable only if potential at infinity is taken as zero.
r
But since the difference in potential is not dependent on this choice
V V kQ
C− =S R 2
In general,
GroundinG concePt
Grounding any conductor gives two informations The potential of the conductor becomes zero.
z
Earth is considered to be an infinite source as well as
z
sink of electrons, hence as many electrons can flow in either direction as required by the conductor to make its potential equal to zero.
MisconcePtion
Grounding a conductor means its charge becomes zero. Well this is true, had the conductor been isolated but not if there were other charged conductors in its vicinity. Let me explain.
Case-I
Isolated grounded conductor (initially charged with Q) : The entire charge flows to earth, + Q
++ + + ++
+ qflown= Q
since if the conductor has any non zero charge, it will create a field which will create a potential difference between
infinity and conductor which is again a contradiction since at infinity also the potential is zero.
Case-II
Concentric shell arrangement : Outer shell with charge Q and inner shell initially uncharged and later grounded. This clearly shows that charge on grounded conductor is non-zero.
Wave oPtics
1. In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths l1 =12000 Å and l2 = 10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit, will a bright fringe from one interference pattern coincide with a bright fringe from the other?
(a) 3 mm (b) 8 mm (c) 6 mm (d) 4 mm 2. Air has refractive index 1.0003. The thickness of air
column, which will have one more wavelength of yellow light (6000 Å) than in the same thickness of vacuum is
(a) 2 mm (b) 2 cm (c) 2 m (d) 2 km 3. The percentage of incident unpolarised light passing
through the two Nicol prisms oriented with their principal planes making an angle q is 25. Then q is (a) 60° (b) 30° (c) 45° (d) 90°
4. White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between slits is b and the screen is at a distance d(>> b) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelengths are
(a) l = b d
b d
2 2
,3 (b) l = b d
b d
2 2
2 3 ,2 (c) l = 2
3 b2
d (d) l = 3
4 b2
5. An unpolarised beam of light is incident on a set d of four polarising plates such that each plate makes an angle of p/3 with preceding sheet. The light transmitted through the combination is
(a) 1/128 (b) 1/256 (c) 1/64 (d) 1/32 6. In a double slit pattern (l = 6000 Å), the 1st order
and 10th order maxima fall at 12.50 mm and 14.75 mm from a particular reference point. If l is changed to 5500 Å, other arrangements remaining the same, position of 10th order maxima will be (a) 12.25 mm (b) 14.55 mm
(c) 15.50 mm (d) 16.55 mm
7. The time period of rotation of the sun is 25 days and its radius is 7 × 108 m. The Doppler shift for the light of wavelength 6000 Å emitted from the surface of the sun will be
(a) 0.04 Å (b) 0.40 Å (c) 4.00 Å (d) 40.0 Å 8. A rocket is going towards moon with a speed v. The
astronaut in the rocket sends signals of frequency u towards moon and receives them back on reflection from the moon. What will be the frequency of signal received by astronaut? (Take v << c)
(a) c c v
u
− (b) c c v
u
− 2 (c) 2v