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Physics Electrostatic Potential and Capacitance

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3 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 F. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ________V.

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Payal Gupta

Contributor-Level 10

$\frac{C V + q}{9 C} + \frac{- 3 C V + q}{3 C} = 0$

$\Rightarrow C V + q - 9 C V + 3 q = 0$

$\Rightarrow q = \frac{8 C V}{4} = 2 C V$

$V_{A B} = \frac{C V + q}{9 C} = \frac{3 C V}{9 C} = \frac{V}{3} = \frac{18}{3} = 6 v o l t$

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3 months ago

Physics Electrostatic Potential and Capacitance Electrostatic Potential and Capacitance

In the given figure, the value of V₀ will be ________ V.

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Vishal Baghel

Contributor-Level 10

K V L Loop 1

6 - l * 1 -l₁ * 1 - 4 = 0

2 = l + l₁ - (i)

K V L Loop 2

4 + l₁ * 1 - (l – l₁) * 1 – 2 = 0

$\Rightarrow 2 + l_{1} - l = 0$

l = 2 + 2l₁ &nb

...more

K V L Loop 1

6 - l * 1 -l₁ * 1 - 4 = 0

2 = l + l₁ - (i)

K V L Loop 2

4 + l₁ * 1 - (l – l₁) * 1 – 2 = 0

$\Rightarrow 2 + l_{1} - l = 0$

l = 2 + 2l₁ - (ii)

from (i) & (ii)

l₁ = 0 v₀ = 4

less

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3 months ago

Physics Electrostatic Potential and Capacitance Class 12th

The equivalent capacitance between points A and B in below shown figure will be…………. $μ F .$

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Payal Gupta

Contributor-Level 10

$\frac{24 * 8}{32} = 6 μ F$

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Physics Electrostatic Potential and Capacitance Class 12th

A long cylindrical volume contains a uniformly distributed charge of density $ρ$ . The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is

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Payal Gupta

Contributor-Level 10

$F_{C} = F_{q}$

$\Rightarrow \frac{m v^{2}}{R} = \frac{ρ R}{2 \in_{0}} q$

$m v^{2} = \frac{ρ R^{2} q}{2 ε_{0}} \Rightarrow \frac{1}{2} m v^{2} = k . E . = \frac{1}{4} \frac{ρ R^{2} q}{\in_{0}}$

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3 months ago

Physics Electrostatic Potential and Capacitance Class 12th

If the charge on a capacitor is increased by 2C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

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Payal Gupta

Contributor-Level 10

$U_{i} = \frac{Q^{2}}{2 C}$

Uf = 1.44 Ui

Qf = Q + 2

$1.44 U_{i} = \frac{{(Q + 2)}^{2}}{2 C}$

$\Rightarrow 1.44 \frac{Q^{2}}{2 C} = \frac{{(Q + 2)}^{2}}{2 C}$

$\Rightarrow 1.2 Q = Q + 2$

0.2 Q = 2

Q = 10 C

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3 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Two identical charged particles each having a mass 10g and charge 2.0 * 10^-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10ms^-2]

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Payal Gupta

Contributor-Level 10

$μ m g = \frac{k q_{1} q_{2}}{L^{2}}$

$\Rightarrow 0.25 * 0.01 * 10 = \frac{9 * 1 0^{9} * 2 * 1 0^{- 7} * 2 * 1 0^{- 7}}{L^{2}}$

$= \frac{9 * 2^{2} * 1 0^{- 5 + 4}}{25 * 10}$

L = 12 cm

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4 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Two point charge Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is:

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alok kumar singh

Contributor-Level 10

According to question, we can write

$F_{C A} = F_{C B} = \frac{K q Q}{x^{2} + {(\frac{d}{2})}^{2}} \Rightarrow F = 2 F_{C A} c o s θ = \frac{2 K q Q x}{{[x^{2} + {(\frac{d}{2})}^{2}]}^{\frac{3}{2}}}$

For maxima of force

$\frac{d F}{d x} = 0, s o$

$x = \frac{d}{2 \sqrt[]{2}}$

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

A positive charge particle of 100mg is thrown in opposite direction to a uniform electric field of strength 1 * 10⁵ NC^-1, If the charge on the particle is $40 μ C$ and the initially velocity is 200 ms^-1, how much distance it will travel before coming to the rest momentarily:

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alok kumar singh

Contributor-Level 10

a = $\frac{q E}{m} = \frac{40 * 1 0^{? 6} * 1 0^{5}}{100 * 1 0^{? 3} * 1 0^{? 3}}$

= 40 * 10³ m/sec²

v² = u² + 2a s

$? 0 = {(200)}^{2} ? 2 * 40 * 1 0^{3} s$

$s = \frac{200 * 200}{2 * 40 * 1 0^{3}} = 0.5 m$

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

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will:

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alok kumar singh

Contributor-Level 10

UI= $\frac{1}{2} c v^{2}$

$U_{i} = \frac{1}{2} \frac{\in_{0} A}{d} * k v^{2}$

$U_{i} = \frac{1}{2} \frac{\in_{0} A}{d} * 10 v^{2}$

$U_{f} = \frac{1}{2} \frac{\in_{0} A}{d} * 15 v^{2}$

$\frac{U_{f}}{U_{i}} = \frac{3}{2}$

$\frac{U_{f} - U_{i}}{U_{i}} * 100 = (\frac{3}{4} - 1) * 100$

$\frac{1}{2} * 100$

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

A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key 'K' is closed, the total energy stored across the combination is E₁. Now key 'K' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E₂. The ratio E₁/E₂ will be:

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alok kumar singh

Contributor-Level 10

I > When switch is closed

Ceq = 2C

Energy E1 = $\frac{1}{2} * C_{e q} * v^{2} = \frac{1}{2} * 2 C * v^{2}$

E1 = CV2

II > When switch is opened

$E_{2} = \frac{1}{2} * (5 c) * V^{2} + \frac{{(C v)}^{2}}{2 * 5 C}$

$= \frac{13}{5} C V^{2}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{5}{13}$

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