Q.466. Electrostatic equilibrium

Question 466.
Three 3.37-g Styrofoam balls of radius 2 cm are coated with carbon black to make them conducting and then are tied to 1.27-m-long threads and suspended freely from a common point. Each ball is given the same charge, q. At equilibrium, the balls form an equilateral triangle with sides of length 24.38 cm in the horizontal plane. Determine the absolute value of q.

Answer 466. Consider any one of the three balls.
Three forces act on it.
i ) its weight, mg, downwards
ii ) tension in the thread, T, in the direction from the ball
     to the point of suspension and
iii ) the resultant of the two forces of repulsion from the remaining two balls.

Let O be the point of suspension, A the position of the ball
and C the center of the equilateral triangle formed by the three balls.

OA = 1.27 m
AC cos30° = 24.38/2 cm = 0.1219 m
=> AC = (0.1219)/cos30° = 0.141 m
=> OC = √[(1.27)^2 - (0.141)^2] m = 1.262 m
=> m∠AOC = arctan[(0.141)/(1.262)] = 6.38°

Vertical component of T balances weight mg
=> Tcos6.38° = 3.37 x 10^-3 x 9.81
=> T = 0.03327 N

Horizontal component of T balances the resultant of the two forces of repulsion
from the remaining two balls
=> 0.03327sin6.38° = 2 * kq^2/(0.2438)^2 * cos30°
=> q^2 = (0.2438)^2 * 0.03327 sin6.38° / (2 * 9 x 10^9 x cos30°)
=> q^2 = 1.409669 x 10^(-14) C
=> q = 0.1187 μC.

Link to YA!

Q.442. Solution of electric network.

Question 442.
What is the current passing through 5 ohm resistor in given circuit ?

Answer 442.
Refer to the figure as under.

Currents, x, y and z and their combinations are shown in different branches.
For the loop ABD,
6x - 2y + 5z = 0 ... ( 1 )
For the loop ABC,
5z + 6(y+z) - 2(x-z) = 0
=> 2x - 6y - 13z = 0 ... ( 2 )

For the loop BCD,
6(y+z) + 2y + 2(x+y) = 10
=> 2x + 10y + 6z = 10 ... ( 3 )

Solving eqns. ( 1 ), ( 2 ) and ( 3 ) using 
Wolfram Alpha Link
z = - 0.4 A
=> current through 5 ohm resistor is 0.4 A.

Link to YA!

Q.417. Electrical resistance / resistivity.

Question 417.
Two cylindrical wires of identical length are made of copper and iron. If they carry the same current and have the same potential difference across their length, what is the ratio of their radii (copper/iron)?

Answer 417.
Same p.d. and same current => same resistance
Resistance of copper wire = ρl/πr^2
and resistance of iron wire = ρ'l/πr'^2
where ρ and ρ' are resiistivity of copper and iron respectively
and r and r' are the radii of the copper and iron wires respectively
As they have the same resistance,
ρl/πr^2 = ρ'l/πr'^2
=> (r/r')
= √(ρ/ρ')
= √(1.68x10^-8)/(10 x 10^-8)
= 0.41

Link to YA!

Q.407. Electrostatic potential

Question 407.
A small object with a mass of 270 mg carries a charge of 35.0 nC and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plates are separated by 2.00 cm. If the thread makes an angle of 14.0° with the vertical, what is the potential difference between the plates?

Answer 407.
Let T = tension in the thread
and F = electrostatic force on the charge
For static equilibrium, balancing forces in the horizontal and vertical directions,
mg = Tcos14° and
F = Tsin14°
Taking the ratio,
F = mgtan14° ... ( 1 )

If V = potential difference between the plates,
electric field between the plates
= V/d = V/0.02 N/C
=> force on the charge
F = 35.0 x 10^-9 V/(0.02) ... ( 2 )
 From ( 1 ) and ( 2 ),
35.0 x 10^-9 V/(0.02) = mgtan14°
=> Potential difference between the plates, V
= (0.02mgtan14°) / (35.0 x 10^-9)
= (0.02 * 270 x 10^-6 * 9.81 * tan14°) / (35.0 x 10^-9) volt
= 377.4 volt.

Link to YA!

Q.353. Maximum electric intensity along the axis of uniformly charged ring

Question 353.
Show that the maximum magnitude, Emax, of the electric field along the axis of a uniformly charged ring occurs at x = a/√2 and has value Q / [6√3 πεo a^2].

Answer 353.
Electric intensity at a distance x from the center of the ring on its axis is given by
E = [Q/(4πεo)] x / (x^2 + a^2)^(3/2)
=> E is maximum where the value of
f (x) = x / (x^2 + a^2)^(3/2) is maximum
For f (x) to be maximum, f '(x) = 0 and f"(x) < 0
ln [f (x)] = lnx - (3/2) ln(x^2 + a^2)
=> f '(x) / f (x) = 1/x - 3x / (x^2 + a^2)
f '(x) = 0
=> 1/x - 3x / (x^2 + a^2) = 0
=> (x^2 + a^2) - 3x^2 = 0
=> x^2 = a^2/2
=> x = ± a/√2
=> E is maximum for x = ± a/√2 and minimum at x = 0
Magnitude of Emax
= [Q/(4πεo)] * (a/√2) / (a^2/2 + a^2)^(3/2)
= [Q/(4πεo)] * (a/√2) / [a^3 * (3/2)^(3/2)]
= [Q/(4πεo)] * (1/a^2) * (1/√2) * (2/3) * √(2/3)
= Q / [6√3 πεo a^2].

Link to YA!

Q.256. To find the equivalent resistance of a circuit of resistors.

Question 256.
Consider an electrical circuit consisting of a cube of 12 identical resistors such that each edge of the cube is a 1 ohm resistor, and each group of three resistors meeting at a vertex are soldered together. Calculate the resistances between nearest neighbor, second-nearest neighbor, and third-nearest neighbor pairs of vertices.

Answer 256.
( I ) Suppose an equivalent resistance is required between adjacent points A and B. (This is the case of the first nearest neighbours) Refer to the figure in the link as under.

Starting from A, the points D and C' are equipotential as they have equivalent paths to B and starting from B, the points D' and C are equipotential as they have equivalent paths to A.
So, sandwiching points D and C' and also D' and C,

(1) A is connected to C' and D by two parallel resistors whose equivalent is 1/2 Ω

(2) At the other end, B is connected to C and D' by two parallel resistors whose equivalent is 1/2 Ω

(3) Between the above two and in series with them is a circuit of resistors as shown in the figure whose equivalent resistance is equal to three parallel resistors, 1 Ω, 2Ω and 1Ω = 2/5 Ω

(4) The above three in series add upto 1/2 + 2/5 + 1/2 = 7/5 Ω

(5) This 7/5 Ω and 1 Ω across A and B are in parallel which equals the final equivalent resistance between
A and B = (7/5) * 1 / (7/5 + 1) = 7/12 Ω.

( II ) Suppose an equivalent resistance is required between points A and D' diametrically opposite on a face of the cube. (This is the case of the second nearest neighbours.)
Refer to the figure in the link as under.

      
(1) C' and B are equipotential points and so are B' and C. Sandwiching them, the circuit will be as shown in figure 2.

(2) Working out equivalent of 5 parallel combinations of resistors, we get the circuit as in figure 3.

(3) In fig. 3, AD and DB' are in series and so are D'A' and A'B' which results in fig. 4.

(4) Fig. 4 is a balanced Wheatstone Bridge. Hence no current flows through the middle 1/2 Ω resistor.

(5) The final circuit is of 1 Ω and 3 Ω resistors in parallel whose equivalent is 3/4 Ω.

(III) Suppose an equivalent resistance is required between diametrically opposite points A and A'.
(This is the case of third nearest neighbour)

AB, AD and AC' are resistors having common point A => B, D and C' are equipotential
A'B', A'D' and A'C are resistors having common point A' => B' D' and C are equipotential
B, D and C' are connected to B', D' and C using six resistors.

Thus, the circuit from A to A' is equivalent to
3 resistors (AB, AD and AC') in parallel connected in series to
6 resistors (BC, BD', DC, DB', C'B', C'D') in parallel connected in series to
3 resistors (B'A', D'A', CA') in parallel
=> equivalent resistance between A and A' is
(1/3) + (1/6) + (1/3) ohm
= 5/6 ohm.

Link to YA!

Q.173. Electrostatics

Question 173.
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ = 1.68 10-7 C/m2, and the plates are separated by a distance of 1.42 10-2 m. How fast is the electron moving just before it reaches the positive plate?

Answer 173.
Electric field between the plates of the capacitor = σ/εo N/C
Force acting on charge q = qσ/εo N
Work done on moving this charge across distance Δx = q(σ/εo)Δx
This work gets converted into kinetic energy of the charge having mass m
=> (1/2) mv^2 = q(σ/εo)Δx
=> v = √[(2q(σ/εo)Δx)/m] ... [As stated by you]

 Now, plugging values,
mass of electron, m = 9.10938188 × 10-31 kg and
charge of electron, q = 1.60217646 × 10-19 C
=> velocity, v
= √[2 * (1.60217646 × 10-19) * (1.68 x 10^-7 / 8.854 x 10-12) * (1.42 x 10^-2 / 9.10938188 × 10-31] m/s
= √[0.94778 x 10^14)
= 9.74 x 10^6 m/s.

Link to YA!

Q.76. Magnetic effects of electric current

Question 76.

A particle of mass m and charge q enters a uniform Magnetic field B (perpendicular to the paper in wards) at P with velocity V at an angle α and leaves the field at Q at an angle β as shown. Explain how these options are correct?

1) α must be equal to β,

2) Length PQ=(2mv*Sin α)/qB,

3) Particle remains in the field for time T=2mα/(qB).

Answer 76.

1)

Since velocity is perpendicular to the direction of magnetic field, the charge experiences force in a direction perpendicular to both. As the force is perpendicular to the velocity, it does no work and results in uniform circular motion of the charge. If the angles α and β are the same, it means that the velcocity of the charge at both these points P and Q is tangential to the circular arc between P and Q inside the magnetic field which is the path of motion of the charge.
2)
The force acting on the charge
= qvB
which equals the centripetal force, making it move on the circular path of radius R given by
qvB = mv^2/R
=> R = mv/qB ... ( 1 )
A perpendicular from the center of the circular path on the chord PQ bisects it at M
PQ
= 2 PM
= 2 Rsinα
Plugging the value of R from ( 1 ) above,
PQ = (2mvsinα)/qB

3)
Arc length PQ
= 2αR
= 2α * mv/(qb)
Time taken to travel this distance with constant velocity, v is
T = distance /velocity
= [2αmv/(qB)]/v
= 2mα/(qB).

LINK to YA!

Q.44. Electrostatics

Question 44.
A charge 4.59 x 10-6 C is held fixed at origin. A second charge 3.40 x 10-6 C is released from rest at the position (1.42 m, 0.746 m). If the mass of the second charge is 6.35 g, what is its speed when it moves infinitely far from the origin?

Answer 44.
Distance between the charges,
r = sqrt ( 1.42^2 + 0.746^2 )
= 1.60 m

Potential energy of the second charge w.r.t. the first charge,
kqq' / r gets fully converted into kinetic energy,
(1/2) mv^2 at an infinite distance.
=> kqq'/r = (1/2) mv^2
=> v = sqrt (2kqq' / ( r*m )
= √[2 (9*10^9) (4.59*10^-6) (3.40*10^-6) / (1.6) (0.00635)
= 3.72 m/s

LINK to YA!

Q.13. Application of Kirchhoff's laws

Question 13.

In the circut below,  R1 and R2 are in parallel and R3 is in series with this combination. There is no immediate place to calculate a voltage drop by Ohm's Law because none of the resistors has both a known current and resistance. Neither can the total resistance be obtained because R2 is unknown. So how can to find the current in R3 by applying Kirchhoff's current Law to node A?

Answer 13.

Let current in R1 = i and in R3 = i'

Applying Kirchoff's first law to node A, i + .001 = i'

Applying Kirchhoff's second law to the outermost loop,

76 = 56000i + 56000 ( i + 0.001)

=> i = 20 /112000 = 0.00018 A = 0.18 mA

=> i' = 1.18 mA

Link to YA!

Q.12. Work done in moving an electric charge in an electric field.

Question 12:

An infinite long plate has surface charge density σ . As shown in the figure, a point charge q is moved from A to B. Net work done by electric field is:

Answer 12:

Electric intensity at any point is constant = σ /2εo and is directed perpendicularly away from the plate.

Net work done

= electric intensity x charge x displacement in the direction of intensity

= (σ /2εo) * q * (x1 - x2).

For theory, CLICK the link of my free educational website.

http://www.schoolnotes4u.com/

Link to YA!

Q.2. Total electrostatic potential energy of a system of charges

Question 2:

Six charges of magnitude +q and -q are fixed at the corners of a regular hexagon of edge length a as shown in the figure.The electrostatic interaction energy of the charged particles is:

Answer 2:

Let the vertices represented by 1, 2, 3, 4, 5 and 6 have charges

+q, - q, +q, - q, + q and - q.

2 out of 6 vertices can be selected in 6C2 ways = 15 ways

=> we have to find PE due to these 15 combination of charges

PE due to charge-combinations, 1-2, 2-3, 3-4, 4-5, 5-6 and 6-1

= - 6q^2 / 4pi*a*εo

PE due to charge combinations, 1-3, 1-5, 2-4, 2-6, 3-5 and 4-6

= + 6q^2 / 4pi*(a√3)*εo

[Note: distance between these charges is a√3]

PE due to charge combinations, 1-4, 2-5 and 3-6

= - 3q^2 / 4pi*(2a)*εo

[Note: distance between these charges is 2a]

Total

= (q^2)/(pi*a*εo) * [- 6/4 + 6/4√3 - 3/8]

= (q^2)/(pi*a*εo) * [√3/2-15/8].

Link to YA!