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Physics Electrostatic Potential and Capacitance

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Two identical metallic spheres A and B when placed at certain distance in air repel each other with force of F. Another identical uncharged sphere C is first place in contact with A and then in contact with B and finally placed at midpoint between sphere A and B. The force experienced by sphere C will be:

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alok kumar singh

Contributor-Level 10

For sphere 'C' -> after contacting with 'A'. q_A = q_C = $\frac{q}{2} .$

...more

For sphere 'C' -> after contacting with 'A'. q_A = q_C = $\frac{q}{2} .$

offer

contacting with 'B'. q_B = q_C = $\frac{3 q}{4}$

$F_{N e t} = | F_{1} - F_{2} |$

F₂ = $\frac{K \cdot 3 q^{2} * 4}{r^{2} 8} = \frac{3}{2} \frac{k q^{2}}{r^{2}}$

$F_{1} = \frac{9 k q^{2} * 4}{16 r^{2}} = \frac{9 k q^{2}}{4 r^{2}} = \frac{9}{4} F$

$∴ F_{N e t} = | \frac{9}{4} F - \frac{3}{2} F |$

= $\frac{9 - 6}{4} F = \frac{3}{4} F$

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

A $60 p F$ capacitor is fully charged by a $20 V$ supply. It is then disconnected from the supply and is connected to another uncharged $60 p F$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $n J$ )

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alok kumar singh

Contributor-Level 10

Common potential after connection.

$V_{common} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{60 * 20 + 0}{120} = 10 V o l t$

Loss of energy $= \frac{1}{2} {C V}^{2} - \frac{1}{2} (2 C) * V_{Common}^{2}$

\begin{array}{r} = \frac{1}{2} * 60 * 10^{- 12} * (20)^{2} - 60 * 10^{- 12} * (10)^{2} \\ = 60 * 10^{- 12} (200 - 100) \\ = 6000 * 10^{- 12} \\ = 6 n J \end{array}

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

A parallel plate capacitor with width 4cm, length 8cm and separation between the plates of 4mm is connected to a battery of 20V. A dielectric slab of dielectric constant 5 having length 1cm, width 4cm and thickness 4mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ___________ $\in_{0} J .$

(Where $\in_{0}$ is the permittivity of free space)

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Vishal Baghel

Contributor-Level 10

$C_{e f f} = [\frac{ε_{0} (7 * 4)}{4 / 10} + \frac{5 ε_{0} (1 * 4)}{4 / 10}] * 1 0^{- 2}$

$C_{e f f} = 1.2 ε_{0}$

Energy = $\frac{1}{2} C_{e f f} V^{2}$ $\frac{}{}_{}^{}$

$\frac{1}{2} (1.2 ε_{0}) (20) (20) = 240 ε_{0}$

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2 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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2 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

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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

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2 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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2 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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2 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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2 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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