Electrostatic Potential and Capacitance

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

A batsman hits back a back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15ms^-1. The impulse imparted to the ball is__________Ns.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Impulse = change in momentum

= $Δ P = 2 m v$

= 2 × 15 × 0.4

= 12 N sec

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

If Electric field intensity of a uniform plane electro magnetic wave is given as $E = - 301.6 s i n (k z - ω t) {\hat{a}}_{x} + 452.4 s i n (k z - ω t) \hat{a} t \frac{V}{m} .$ 

Then, magnetic intensity ‘H’ of this wave in Am^-¹ will be:

[Given : Speed of light in vacuum c = 3 × 10⁸ ms^-¹, Permeability of vacuum $μ_{0} = 4 π \times 1 0^{- 7} N A^{- 2}$ ]

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Since

$\vec{E} . \vec{B} = 0 (\vec{E} ⊥ \vec{B})$

$& \frac{E_{0}}{B_{0}} = C = 3 \times 1 0^{8} m / s e c$

for option D

$\frac{E_{0}}{B_{0}} = \frac{\sqrt[]{301 . 6^{2} + 452 . 4^{2}}}{1 0^{- 6} \sqrt[]{1 . 5^{2} + 1^{2}}} \begin{matrix} = 301.6 \times 1 0^{6} & = 3.01 \times 1 0^{8} m / s e c \end{matrix}$

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

A parallel plate capacitor with plate area A and plate separation d = 2m has a capacitance of $4 μ F .$ The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K = 3 (as shown in figure) will be:

31 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$G i v e n \frac{\in_{0} A}{d} = 4 μ F$

$C_{1} = \frac{\in_{0} A \times 2}{d} \begin{matrix} = 2 \times 4 & = 8 μ F \end{matrix}$

$C_{2} = \frac{k \in_{0} A}{d / 2} = 3 \times 2 \times 4 = 24 μ F$

$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{8 \times 24}{8 + 24} = \frac{8 \times 24}{32} = 6 μ F$

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

The change on capacitor of capacitance $15 μ F$ in the figure given below is:

10 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} = \frac{1}{10} + \frac{1}{15} + \frac{1}{20}$

$\frac{1}{C_{e q}} = \frac{6 + 4 + 3}{60} = \frac{13}{60}$

$L_{e q} = \frac{60}{30} μ F$

$Q = C_{e q} v = \frac{60}{13} \times 13 μ C$

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

A person is standing in an elevator. In which situation, he experiences weight loss?

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

F.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss.

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

A composite parallel plate capacitor is made up to two different dielectric materials with different thickness (t₁ and t₂) as shown in figure. The two different dielectric materials are separated by a conducting foil F. The voltage of the conducing foil is ___________V.

16 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Capacitance of each capacitor

$C_{1} = \frac{A 3 ε_{0}}{\frac{1}{2}} = 6 A ε_{0}$

$C_{2} = A 4 ε_{0} = 4 A ε_{0}$

Equivalent capacitance

$Δ v_{2} = \frac{240 A ε_{0}}{4 A ε_{0}} = 60 v$

$v_{f o i l} = 60 v$

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is……… times that of a smaller drop.

9 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

With the help of conservation of volume, we can write

$27 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 3 r . . . . . . . (1)$

With the help of conservation of charge, we can write

Q = 27 q. (2)

Potential energy of single drop = U1 = $\frac{q^{2}}{8 π ε_{0} r}$

Potential energy of bigger drop = $U_{2} = \frac{Q^{2}}{8 π ε_{0} R} = \frac{27 \times 27 \times q^{2}}{8 π ε_{0} (3 r)} = 243 (\frac{q^{2}}{8 π ε_{0} r}) = 243 U_{1}$