On introduction of dielectric slab to fill the gap between plates of capacitor completely when capacitor is connected with battery.
(i) The potential difference V between capacitors is same due to connectivity with battery and hence, charge q' becomes K times of original charge as
q' = C'V' = (KC) (V) = K(CV) = Kq
q' = Kq
(ii) Electric field intensity continue to be the same as potential difference and separation between two plates remain unaffected as
\(E=\frac { V }{ d } \)
(iii) The capacitance of capacitor becomes K times of original capacitor.
\(\therefore \ C'=KC=\frac { K{ \varepsilon }_{ 0 }A }{ d } \) 

(i) Force acting on the particle, F = Bqv
In vector form, F = q(v x B)
where, B is uniform magnetic field and v is velocity with particle which is moving.
From this equation, it is clear that direction of force is perpendicular to the plane containing both v and B. In other words, force acts perpendicular to both v and B. When velocity becomes perpendicular to force, the path of the object becomes circular.

In this case, B is assumed to act perpendicular to v. In case, B is not perpendicular to v, a component of v remains perpendicular to v. It creates circular path. The component of v parallel to B will create linear path. Here, circular path is due to v cosθ and linear path is due to v sinθ. Both when combined gives helical path.
(ii) Since, force always adjusts itself in a direction which becomes perpendicular to velocity, so only direction of velocity is changes not the magnitude. Hence, the kinetic energy of the, particle always remains constant.

Since, displacement current = conduction current.
Therefore, 
I_d = \(\frac { V }{ X_{ c } } \) = \(\frac { 220 }{ 100 } \)
\(\Rightarrow \) I_d = 2.2A

\(B=1 \ Wb{ m }^{ -2 }; \ n=80; \ A=0.01{ m }^{ 2 }; \ dt=0.1s; \ e=?\)
\(e=\frac { d\phi }{ dt } =\frac { nAB }{ dt } \)
or   \(e=\frac { 80\times 0.01\times 1 }{ 0.1 } =8V.\)

An ideal inductor is a coil consisting of large number of turns wound in such a way that magnetic flux linked with it due to current passing through it is confined to a small region only. The ohmic resistance of such a coil is zero.

Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T₂ = 100°C
Resistance of the silver wire at T₂, R₂ = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as 
\(\alpha=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}\)
\(=\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1}\)
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.

Here, \({4\pi \epsilon_oab\over b-a}=4\pi\epsilon_oR\)
\({ab\over b-a}=R\)
As b - a = 2 cm  R = \({1.2\over 2}m=60cm\)
\({ab\over 2}=60 \ or \ ab=120\)
As \((b+a)^2=(b-a)^2+4ab=(2)^2+4\times 120=484\)
b + a = 22
2 + a + a = 22
a = 10 cm, b = 12 cm

\(Here,\ I=5sin\ 100\pi t\)
\( Compare\ with\ I={ I }_{ 0 }sin\ \omega t\)
\({ I }_{ 0 }=5A,\ \omega =2\pi v=100\pi\)
\(v=\frac { 100\pi }{ 2\pi } =50 \ Hz\)
\(\ { I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 5 }{ \sqrt { 2 } } =\frac { 5\sqrt { 2 } }{ 2 } =3.54A\)

It states that current flowing through a conductor is proportional to the potential difference across its two ends provided the physical conditions of the conductor remain unchanged.
If V is the potential difference between two ends of a conductor and I is the current flowing through it, then
\(V\alpha I\)
\(V=RI\)
Where R is the constant of proportionality and is called the resistance of the conductor. Its value depends upon
(i) Shape of the conductor
(ii) Length of conductor
(iii) Nature of the material

If a graph is plotted between V and I, the graph will be a straight line passing through the origin.
(b) Resistance is the property of a material by virtue of which it opposes the flow of current through it and quantitatively it is given by,
\(R=\frac { V }{ I }\)
\( =\frac { Potential \ difference }{ Current } \)
definition of ohm
Hence a conductor has a resistance of one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its two ends.

Same as velocity of light

\(Pα 1/R (i.e) \)25 watts

No, the electric field due to an infinite plane sheet of charge does not depend upon the distance of the observation point from the plane sheet of charge.

Kilowatt is the unit of power. 1 kilowatt = 100 watt. Kilowatt hour is the unit of electric energy, where
1 kilowatt hour = 1000 watt x 1 hour
= 1000 watt x 60 x 60 second
= 3.6 x 10⁶ J

Whereas in electricity, an isolated charge exists; in magnetism, an isolated magnetic pole does not exist.

The S.I unit of electric potential is volt. The electric potential at a point is said to be 1 volt when 1J of work is done in moving a positive charge of 1 coulomb from infinity to that point against the electrostatic force.

oscillating electric and magnetic field of an electromagnetic wave are in the same phase

Induced emf is called back emf as it opposes the growth as well as decay of current in the circuit.