Electric Charges and Fields

1.32 (a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss's law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.

(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null-point. When a test charge is displaced along the line, it experiences a restoring force, If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.

1.31 A proton has 3 quarks. Let there be n 'up quarks', then number of 'down quarks' = 3-n

Charge due to n 'up quarks' = $(\frac{2}{3}$ e) n

Charge due to (3-n) 'down quarks' =$-$$\frac{1}{3}e\left)\right(3-n)$ 

Total charge on a proton = +e = $(\frac{2}{3}$ e) n +  ( -$\frac{1}{3}e\left)\right(3-n)$ 

e =  $\frac{2ne}{3}$ +$\frac{ne}{3}$-e

2e = $\frac{3ne}{3}$

n = 2

Number of 'up quark' = 2 and number of 'down quark' = 1. Therefore a proton can be represented as 'uud'.

A neutron has 3 quarks. Let there be n 'up quark' in a neutron and (3-n) 'down quark'

Charge due to n 'up quark' = +$(\frac{2}{3}$ e)n

Charge due to (3-n) 'down quark' = 
- (

Since total charge of a neutron is zero, we get

+ $(\frac{2}{3}$e)n -($\frac{1}{3}e\left)\right(3-n)$ = 0

$(\frac{2}{3}$ e)n = ( $\frac{1}{3}e\left)\right(3-n)$

en = e or n = 1

Hence number of 'up quark' in neutron = 1

Number of 'down quark' in neutron = 3-1 = 2

Therefore a neutron can be represented as 'udd'.

1.30 

Take a long thin wire XY of uniform linear charge density $\lambda$. Consider a point A at a perpendicular distance l from the midpoint O of the wire. Let E be the electric field at point A due to the wire XY. Consider a small length element dx on the wire section with OZ = x. Let q be the charge on this piece.

Q = $\lambda dx$

Electric field due to the piece,

dE =  = $\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda dx}{{\left(AZ\right)}^2}$  = $\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda dx}{(l^2+x^2)}$since AZ =$\text{exception 25:}$

The electric field is resolved into two rectangular components. dE $\cos ?\theta$ is the perpendicular component and dE$\sin ?\theta$ $$ When the whole wire is considered, the component dE $\sin ?\theta \mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{e}\mathrm{l}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}.\mathrm{}$ is cancelled. Only the perpendicular component dE $\cos ?\theta$ affects point A.

Hence, effective electric field at point A due to element dx is$d{E}_1$ 

 ………………….(1)

In  , $?AZO$ ,  $\tan ?\theta =\frac{x}{l}$ ,x =$l\tan ?\theta$……….(2)

On differentiating equation (2), we get

$\frac{dx}{d\theta }=l{sec}^2\theta$ , dx = …………….(3)

From equation (2), we have

 $x^2$ +$l^2$ = $l^2{tan}^2\theta +l^2$ =$l^2{sec}^2\theta$…….(4)

Putting equations (3) and (4) in equation (1), we get

 $d{E}_1=\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda l{sec}^2\theta d\theta \cos ?\theta }{l^2{sec}^2\theta }$ =$\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda \cos ?\theta d\theta }{l}$…….(5)

The wire is so long that $\theta$tends from  $-\frac{\pi }{2}$to$\frac{\pi }{2}$

By integrating equation (5), we get

 ${\int }_{-\frac{\pi }{2}}^{\mathrm{}\frac{\pi }{2}}d{E}_{\mathrm{1=}}$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda \cos ?\theta d\theta }{l})$

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{l}{\left[\sin ?\theta \right]}_{\mathrm{}-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{l}*2=\frac{\lambda }{2\pi {\epsilon }_{0}l}$

Therefore the electric field due to long wire is $\frac{\lambda }{2\pi {\epsilon }_{0}l}$

1.29 Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let E be the electric field outside the conductor, q is the electric charge, $\sigma$is the charge density and ${\epsilon }_0$ is the permittivity of free space.

Charge q =$\sigma$ $*ds$ 

According to Gauss's law, flux$\phi$ = E.ds = $\frac{q}{{?}_0}$=$\frac{\sigma \mathrm{}*ds}{{?}_0}$

Hence, E = $\frac{\sigma }{2{\epsilon }_0}\stackrel{}{n}$

Therefore, the electric field just outside the conductor is $\frac{\sigma }{{2\epsilon }_0}\stackrel{}{n}$ . This field is a superposition of field due to the cavity E' and the field due to the rest of the charged conductor E'. These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor.

So E' + E' = E

E'  = $\frac{E}{2}$=$\frac{\sigma }{2{\epsilon }_0}\stackrel{}{n}$

Hence, the field due to the rest of the conductor is $\frac{\sigma }{2{\epsilon }_0}\stackrel{}{n}$
$\frac{}{}\stackrel{}{?}$

1.28 (a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.

Let q be the charge inside the conductor and ${?}_0$ the permittivity of free space.

According to Gauss's law, Flux,$\phi$  = E.ds = $\frac{q}{{?}_0}$

Here, E = 0, hence $\frac{q}{{?}_0}$ = 0, so q = 0 (as ${?}_0\ne 0)$

Therefore, the charge inside the conductor is zero. The entire charge Q appears on the outer surface of the conductor.

(b) The outer surface of the conductor A has a charge amount Q. Another conductor B, having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount –q will be induced in the inner surface of the conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of the conductor A is Q+q.

(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.

Ans.1.26

(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.

(b) The field lines shown in (b) do not represent electrostatic field lines because field lines can not emerge from a negative charge and cannot terminate at a positive charge.

(c) The field lines shown in (c) represent electrostatic field line. This is because the field lines emerge from the positive charge and repel each other.

(d) The field lines shown in (d) do not represent electrostatic field lines because the field lines should not intersect each other.

(e) The field lines shown in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.