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Electrostatic Potential and Capacitance

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2 months 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.

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Vishal Baghel

Contributor-Level 10

Impulse = change in momentum

= $Δ P = 2 m v$

= 2 * 15 * 0.4

= 12 N sec

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2 months 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 $μ_{0} = 4 π * 1 0^{- 7} N A^{- 2}$ ]

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Payal Gupta

Contributor-Level 10

Since

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

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

for option D

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

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2 months 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:

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Vishal Baghel

Contributor-Level 10

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

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

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

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

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2 months 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:

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Vishal Baghel

Contributor-Level 10

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

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

$L_{e q} = \frac{60}{30} μ F$

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

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2 months 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?

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Payal Gupta

Contributor-Level 10

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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2 months 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.

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Vishal Baghel

Contributor-Level 10

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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2 months 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.

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Payal Gupta

Contributor-Level 10

With the help of conservation of volume, we can write

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

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 = $U_{2} = \frac{Q^{2}}{8 π ε_{0} R} = \frac{27 * 27 * q^{2}}{8 π ε_{0} (3 r)} = 243 (\frac{q^{2}}{8 π ε_{0} r}) = 243 U_{1}$

$\Rightarrow \frac{U_{2}}{U_{1}} = 243$

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2 months ago

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

The total charge on the system of capacitors $C_{1} = 1 μ F, C_{2} = 2 μ F, C_{3} = 4 μ F a n d C_{4} = 3 μ F$ connected in parallel is:

(Assume a battery of 20V is connected to the combination)

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Vishal Baghel

Contributor-Level 10

$C_{e q} = C_{1} + C_{2} + C_{3} + C_{4}$

= 1 + 2 + 4 + 3

$= 10 ? F$

Q = C_eq v

= (10 x 20) $? C$

Q = 200 $? C$

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2 months ago

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

If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of $λ$ where C/V = $λ ?$

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Payal Gupta

Contributor-Level 10

$λ = \frac{c}{v} = \frac{Q^{3}}{W^{2}} = \frac{I^{3} T^{3}}{M^{2} L^{4} T^{- 4}} \Rightarrow [λ] = [M^{- 2} L^{- 4} T^{7} l^{3}]$

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3 months ago

Electrostatic Potential and Capacitance Class 11th

Given below are two statements.

Statement I : Electric potential is constant within and at the surface of each conductor.

Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.

In the light of the above statements, choose the most appropriate answer from the options given below.

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Vishal Baghel

Contributor-Level 10

'A' at static equilibrium, E_{(inside conductor)}= 0

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