Question

Q3. An electric dipole of moment p is placed parallel to the uniform electric field. The amount of work done in rotating the dipole by 900 is-
(i) 2pE
(ii) pE
(iii) pE/2
(iv) Zero

Q4. Three capacitors 2µF, 3µF and 6µF are joined in series with each other. The equivalent capacitance is-
(i) 1/2µF
(ii) 1µF
(iii) 2µF
(iv) 11µF

Q5. Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be-
(i) 5F
(ii) F
(iii) F/2
(iv) F/5

Q6. Which statement is true for Gauss law-
(i) All the charges whether inside or outside the gaussian surface contribute to the electric flux.
(ii) Electric flux depends upon the geometry of the gaussian surface.
(iii) Gauss theorem can be applied to non-uniform electric field.
(iv) The electric field over the gaussian surface remains continuous and uniform at every point.

Q7.A capacitor plates are charged by a battery with ‘V’ volts. After charging battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become
(i)V/2
(ii) V/K
(iii)KV
(iv) Zero​

Answer 1

Q3. (ii) pE

W = pE (cosθ1 – cosθ2)

θ1 = 0°

θ2 = 90°

W = pE (cos 0° - cos 90°) = pE (1 – 0) = pE

Q4.(ii) 1µF

Given:- C1=2µF , C2=3µF, C3=6µF

When capacitors are in series total capacitance found by adding the reciprocals of the individual capacitors.

Total capacitance  $=\frac{1}{2} +\frac{1}{3}+\frac{1}{6}$

= 1 µF

Q5. (i) 5F

Given:-   K=5

let two charge Q₁ and Q₂ are placed in a medium of dielectric constant, K = 5, distance between them is r.

from Coulomb's law, force between charges will be , F =  ...(1)

now the same charges are placed in vacuum and also the seperation between them is same.

force between them will be , F' =  ...(2)

diving equation (2) by equation (1) we get,

⇒ F'/F = K/1

⇒ F' = KF = 5F [∵ K = 5 ]

Q6. (iv) The electric field over the gaussian surface remains continuous and uniform at every point.

Gauss law states that electric flux through a closed surface = total enclosed charge divided by electrical permittivity of vacuum.

Q7. Q = Charge remains constant Cʹ = K C

Qʹ = Cʹ Vʹ

Q = Cʹ Vʹ

Q = K C Vʹ

Vʹ =  Q/ K C

Vʹ= V K

Answer 2

Q3. The amount of work done in rotating the dipole is :

$W = PE \{1 - \cos( \theta) \}$

$\implies W = PE \{1 - \cos( {90}^{ \circ} ) \}$

$\implies W = PE \{1 - 0 \}$

$\implies W = PE$

Q4. The equivalent capacitance (since the capacitors are in series) is :

$\dfrac{1}{C_{eq}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}$

$\implies\dfrac{1}{C_{eq}} = \dfrac{3 + 2 + 1}{6}$

$\implies\dfrac{1}{C_{eq}} = \dfrac{6}{6}$

$\implies C_{eq} = 1 \: \mu F$

Q5. When dielectric slab of dielectric constant = 5 is inserted in between two charges, the new force will becomes 5 times the initial force.

$F_{new} = 5F$

Hope It Helps.