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 t

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

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 o

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 =$\sigma$ $\times ds$ 

According to Gauss’s law, flux$\phi$ = E.

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,$\phi$  =

1.27 Dipole moment of the system, p = q$\times dl$ = ${10}^{-7}$ Cm

Rate of increase of electric field per unit length,$\frac{dE}{dl}$ =${10}^5$ N$C^{-1}$ 

Force experienced by the system is given by the relation, F = qE = q$\times \frac{dE}{dl}\times dl$ 

= (q  $\times dl)\times \frac{dE}{dl}$ =p $\times \frac{dE}{dl}$ =$-{10}^{-7}\times$ ${10}^5$= ${-10}^{-2}$ N

The force is  N in the negative z-direction i.e. opposite to the direction of

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 c

1.25 Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 $\times {10}^{4}N{C}^{-1}$$\mathit{}$

Density of oil, $\rho =$ 1.26 g / ${cm}^3$$$ = 1.26$\times {10}^3$  g/ $m^3$

Acceleration due to gravity, g = 9.81 m/$s^2$ 

Charge of an electron, e = 1.60$\times {10}^{-19}$C

Let the radius of the oil drop be r

Force (F) due to electric field (E) is equal to the wei

1.24 The given conditions are explained in the adjacent diagram

Where A and B represent two large, thin metal plates, parallel and close to each other. The outer surface of A is shown as I, outer surface of B is shown as II and the surface in between A and B is shown as III.

Charge density of pl

1.23 Electric field produced by the infinite line charge at a distance d having linear charge density$\lambda$ 
 is given by

E = $\frac{\lambda }{2\pi {\epsilon }_{0}d}$, where

E = Electric field = 9 $\times {10}^4$ N/C

${?}_0$ = Permittivity of free space = 8.854  $\times {10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$

d = 2 cm = 0.02 m

Hence, $\lambda$ = 9  $\times {10}^4$ $\times 2\times \pi \times$8.854 $\times {10}^{-12}\times$
 0.02 = 10$\mu C/m$