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Physics Electrostatic Potential and Capacitance

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3 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Three capacitors C₁ = $2 μ F, C_{2} = 12 μ F$ and C₃ = $12 μ F$ are connected as shown in figure. Find the ratio of the charges on capacitors C₁, C₂ and C₃ respectively:

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alok kumar singh

Contributor-Level 10

$Q_{1} = C_{1} V = 2 V μ C$

$Q_{2} = Q_{3} = \frac{C_{2} C_{3}}{C_{2} + C_{3}} V = \frac{6 * 12}{6 + 12} V = 4 V μ C$

$Q_{1} : Q_{2} : Q_{3} = 1 : 2 : 2$

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Physics Electrostatic Potential and Capacitance Class 12th

27 identical drops are charged at 22V each. They combine to form a bigger drop. The potential of the bigger drop will be…………V.

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alok kumar singh

Contributor-Level 10

Volume of 27 identical drops = volume of a bigger drop

$\Rightarrow 27 * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$

R³ = 27r³

R = 3r

Given potential of a small drop = 22v

$V_{b i g g e r} = \frac{k (27 q)}{R} = \frac{27 k q}{3 r} = 9 \frac{k q}{r} = 9 * 22 = 198 v o l t$

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Physics Electrostatic Potential and Capacitance Class 12th

Two metallic plates form parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness and of area equal to area of each plate is introduced between the plates. What will be ratio of the new capacitance to the original capacitance of the capacitor?

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alok kumar singh

Contributor-Level 10

$Δ v = \int \vec{E} . d \vec{x}$

Δ v = \frac{σ}{\in o} \frac{d}{2}

C_{N e w} = \frac{Q}{Δ v} = \frac{σ A 2 \in o}{σ d} = \frac{2 \in o A}{d}

$C_{N e w} = 2 {\begin{array}{l} C \\ o r i g i n a l \end{array}}_{}$

$\frac{C_{N e w}}{C_{O r i g i n a l}} = 2 : 1$

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Physics Electrostatic Potential and Capacitance Class 12th

Two identical positive charge Q each are fixed at a distance of '2a' apart from each other. Another point charge q₀ with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge q₀ executes SHM. The time period of oscillation of charge q₀ will be:

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alok kumar singh

Contributor-Level 10

$F_{R} = {\frac{k Q q_{0}}{{(a - x)}^{2}} - \frac{k Q q_{0}}{{(a + x)}^{2}}}$

$= - k Q q_{0} {\frac{a^{2} + x^{2} + 2 a x - a^{2} - x^{2} + 2 a x}{(a^{2} - x^{2})}}$

$F_{R} = - \frac{k Q q_{0} (4 a x)}{{(a^{2} - x^{2})}^{2}}$

$T = 2 π \sqrt[]{\frac{a^{3} m}{4 k Q q_{0}} = \sqrt[]{\frac{4 π^{2} a^{3} m * 4 π ε_{0}}{4 Q q_{0}}}} = \sqrt[]{\frac{4 π^{3} ε_{0} m a^{3}}{Q q_{0}}}$

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3 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Capacitance of an isolated conduction sphere of radius R₁ becomes n times when it is enclosed by a concentric conducting sphere of radius R₂ connected to earth. The ration of their radii $(\frac{R_{2}}{R_{1}})$ is

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Payal Gupta

Contributor-Level 10

$C_{1} = 4 π R_{1} a n d$

$C_{2} = \frac{4 π R_{2} R_{1}}{R_{2} - R_{1}} = \frac{R_{2} C_{1}}{R_{2} - R_{1}}$

$\Rightarrow \frac{C_{2}}{C_{1}} = \frac{4 π R_{2} R_{1}}{R_{2} - R_{1}} = \frac{R_{2}}{R_{2} - R_{1}} = n$

$\Rightarrow \frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$

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3 months ago

Physics Electrostatic Potential and Capacitance Electrostatic Potential and Capacitance

Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of constant K such that the equivalence capacitance of the system became $24 μ F$ . The value of K will be:

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Vishal Baghel

Contributor-Level 10

$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{40 * 40 k}{40 (1 + k)} = 24$

$\begin{array}{l} \Rightarrow 40 k = 24 + 24 k \\ \Rightarrow 16 k = 24 \end{array}$

$k = \frac{24}{16} = \frac{3}{2} = 1.5$

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3 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Two identical thin metal plates has charge q₁ and q₂ respectively such that q₁ > q₂. The plates were brought close to each other to from a parallel plate capacitor of capacitance C. The potential difference between them is:

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alok kumar singh

Contributor-Level 10

V = $\frac{q_{(c a p a c i t a n c e)}}{C} = \frac{q_{1} - q_{2}}{2 c}$

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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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3 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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3 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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