Numerical - 7

Q . The electric field components in Fig. 1.27 are E(x)  = αx 1/2 , E(y) = E(z)  = 0, in which α = 800 N/C m1/2. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m.

Solution :

(a) Since the electric field has only an x component, for faces

perpendicular to x direction, the angle between E and ΔS is ± π/2. 

Therefore, the flux φ = E.ΔS is separately zero for each face

of the cube except the two shaded ones. 

Now the magnitude of the electric field at the left face is E(L)  = αx 1/2 = αa 1/2 (x = a at the left face).

The magnitude of electric field at the right face is

E(R) = α x 1/2 = α (2a) 1/2    (x = 2a at the right face).

The corresponding fluxes are

φ(L) = E(L) .ΔS = ΔS E(L) . n^ (L) =E(L) ΔS cosθ = –E(L) ΔS, 

since θ = 180°

φ(L)= –E(L) a^2

φ(R) = E(R).ΔS = E(R) ΔS cosθ = E(R) ΔS, since θ = 0°

φ(R)= E(R) a^2

Net flux through the cube = φ(R) + φ(L)  = E(R)a^2 – E(L) a^2 

= a^2 [E(R)  – E(L)] = αa^2 [(2a) ^1/2 – a ^1/2]

= αa ^5/2 [2^1/2 - 1] = 1.05 N m2 C–1

(b) We can use Gauss’s law to find the total charge q inside the cube.

We have φ = q/ε0 or q = φε 0 . Therefore,

q = 1.05 × 8.854 × 10–^12 C = 9.27 × 10^–12 C

Numerical-4

Q .  Two point charges q1 and q2 , of magnitude +10^–8 C and 

–10^–8 C, respectively, are placed 0.1 m apart. Calculate the electric

fields at points A, B and C shown in Fig. 1.14.

Solution : 

The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude

E(1A) =  (9 x 10 Nm 2 C-2 ) x (10^8 C) / (0.05 m)^2  = 3.6 × 10^4 N C–1 

The electric field vector E(2A) at A due to the negative charge q2 points towards the right and has the same magnitude. 

Hence the magnitude of the total electric field E(A)  at A is

E(A)  = E(1A) + E(2A) = 7.2 × 10^4 N C–1

E(A)  is directed toward the right.

The electric field vector E(1B) at B due to the positive charge q1 points towards the left and has a magnitude

E(1B) =  (9 x 10 Nm 2 C-2 ) x (10^8 C) / (0.05 m)^2  = 3.6 × 10^4 N C–1 

The electric field vector E(2B) at B due to the negative charge q2 points towards the right and has a magnitude

E(2B) =  (9 x 10 Nm 2 C-2 ) x (10^8 C) / (0.15 m)^2  = 4 × 10^3 N C–1 

The magnitude of the total electric field at B is

E(B) = E(1B) – E(2B) = 3.2 × 10^4 N C–1

E(B)  is directed towards the left.

The magnitude of each electric field vector at point C, due to charge

q1 and q2 is

E(1C) - E(2C)=  (9 x10 Nm 2 C-2 ) x (10^8 C) / (0.10 m)^2  = 9 × 10^3 N C–1 

The directions in which these two vectors point are indicated in

Fig. 1.14. The resultant of these two vectors is

E(C) = E(1) cos π/3 + E(2) cos π/3 = 9 × 10^3 N C–1

E(C)  points towards the right.

 

Numerical-3

Q.  Consider three charges q1 , q2 , q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.9? 

Solution 

In the given equilateral triangle ABC of sides of length l, if

we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3/2 ) l and the distance AO of the centroid O from A is (2/3) AD = (1/ 3 ) l. By symmatry AO = BO = CO.

Thus, Force F1 on Q due to charge q at A = 3Qq / 4πεo l^2   along AO

 Force F2 on Q due to charge q at B = 3Qq / 4πεo l^2   along BO

 Force F3 on Q due to charge q at C = 3Qq / 4πεo l^2   along CO

The resultant of forces F2 and F3 is  3Qq / 4πεo l^2   along AO

by the parallelogram law. 

Therefore, the total force on Q = 3Qq / 4πεo l^2 (r^ - r^) = 0

where rˆ is the unit vector along OA

It is clear also by symmetry that the three forces will sum to zero.

Suppose that the resultant force was non-zero but in some direction.

Consider what would happen if the system was rotated through 60º

about O.

Numerical-4

Q. Consider the charges q, q, and –q placed at the vertices

of an equilateral triangle, as shown in Fig. 1.10. What is the force on

each charge? 

Solution 

The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in Fig. 1.10. 

By the parallelogram law, the total force F1 on the charge q at A is given by

F1 = F r^1

where r^1 is a unit vector along BC.

The force of attraction or repulsion for each pair of charges has the

same magnitude 

F = q^2 / 4πεo l^2 

The total force F2 on charge q at B is thus F2 = F r^2, where r^2 is a unit vector along AC. 

Similarly the total force on charge –q at C is F3 = 3 F n^

where n^ is the unit vector along the direction bisecting the ∠BCA.

It is interesting to see that the sum of the forces on the three charges is zero, i.e.,

F1 + F2 + F3 = 0 

Numerical-1

Q . If 10^9 electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?

Solution :

In one second 10^9 electrons move out of the body. Therefore

the charge given out in one second is 1.6 × 10^–19 × 10^9 C = 1.6 × 10–^10 C.

The time required to accumulate a charge of 1 C can then be estimated to be 1 C ÷ (1.6 × 10^–10 C/s) = 6.25 × 10^9 s = 6.25 × 10^9 ÷ (365 × 24 × 3600) years = 198 years. 

Thus to collect a charge of one coulomb, from a body from which 10^9 electrons move out every second, we will need approximately 200 years. 

One coulomb is, therefore, a very large unit for many practical purposes.

It is, however, also important to know what is roughly the number of

electrons contained in a piece of one cubic centimetre of a material.

A cubic piece of copper of side 1 cm contains about 2.5 × 10^24

electrons.

Numerical-2

Q.  How much positive and negative charge is there in a cup of water?

Solution : 

Let us assume that the mass of one cup of water is 250 g. The molecular mass of water is 18g. Thus, one mole (= 6.02 × 10^23 molecules) of water is 18 g. 

Therefore the number of molecules in one cup of water is (250/18) × 6.02 × 10^23.

Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. 

It is equal to

(250/18) × 6.02 × 10^23 × 10 × 1.6 × 10^–19 C = 1.34 × 10^7 C.