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Physics Electrostatic Potential and Capacitance

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

An electron moves in the field produced by a charged sphere of radius 10cm along the radius connecting the points separated by 12 and 15 cm from the centre of the sphere. The velocity of the electron changes thereby from 2 * 10⁵ to 2 * 10⁶ m/s. Determine the surface charge (in nC/m²) density of the sphere. (Mass of electron = 9.1 * 10^-31kg)

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Vishal Baghel

Contributor-Level 10

$\frac{k Q e}{0.12} + \frac{1}{2} m u^{2} = \frac{k Q e}{0.15} + \frac{1}{2} m u^{2}$

$\frac{k Q e * 0.03}{0.12 * 0.15} = \frac{1}{2} m * (v^{2} - u^{2})$

$Q = \frac{9.1 * 99 * 1 0^{- 11}}{12}$

$= \frac{75.075 * 1 0^{- 12}}{4 π * {(0.1)}^{2}} = 5.97 n C / m^{2}$

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Two air filled capacitors are connected as shown. The left capacitor is suddenly filled with a dielectric of constant K. The work done by battery after that is $24 μ J .$ Dielectric constant K is:-

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Vishal Baghel

Contributor-Level 10

$W D = Δ Q V$

$\Rightarrow Δ Q = 4 μ C$

$Q_{i} = \frac{2 * 4}{2 + 4} * 6 = 8 μ C$

$Q_{f} = 12 μ C = C_{e q} * 6$

$C_{e q} = 2 μ F \Rightarrow C_{1} = C_{2} = 4 μ F$

$k = \frac{C_{1}}{C_{0}} = 2$

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2 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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Vishal Baghel

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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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

A capacitor is connected to a 20 V battery through a resistance of 10 $Ω .$ It is found that the potential difference across the capacitor rises to 2V in 1 $μ s .$ The capacitance of the capacitor is________ $μ F .$

Given $I n (\frac{10}{9}) = 0.105$

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Vishal Baghel

Contributor-Level 10

V = 10 $(1 ? e^{? t / R C})$ $^{}$

2 = 20 $(1 ? e^{? t / R C})$ $^{}$

$\frac{1}{10} = 1 ? e^{? t / R C}$

$e^{t / R C} = \frac{10}{9}$

$\frac{t}{R C} = l n (\frac{10}{9}) = 0.105$

$C = \frac{t}{R * 0.105} = \frac{1 0^{? 6}}{10 * 0.105} = 0.95 ? F$

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

A capacitor of capacitance 3µF is charged to a potential difference of 12V. Another identical capacitor, initially uncharged is filled with a dielectric having dielectric constant K and it is connected to the charged capacitor. The final common potential difference across both capacitors is found to be 2V. Find the value of k.

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Vishal Baghel

Contributor-Level 10

V_common = (C? V? + C? V? )/ (C? + C? )? 2 = (3? F)*12)/ (3? F)+ (3? F)*K)? 2 = 12/ (K+1)? K = 5

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

A proton and an $α$ -particle both initially at rest are accelerated in a region of electric potential difference V and gravity free space. The proton is projected with gained kinetic energy against a uniform constant electric field and comes to momentary rest after travelling a distance $S_{0}$ . If the $α$ -particle is also projected with gained kinetic energy against same electric field, it will come to momentary rest after travelling through:

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Raj Pandey

Contributor-Level 9

Gain in K.E. $=$ Loss in P.E.

$_{}$ $K_{p} = e V$ for proton

$K_{α} = 2 e V$ for $α$ - particle

Again, Loss in K.E. $=$ Work against field

$K_{p} = e E \cdot S_{0}$ for proton

$\Rightarrow e V = e E S_{0}$

$∴ S_{0} = \frac{V}{E}$

$K_{α} = 2 e E . S$ for $α -$ particle

$\Rightarrow 2 e V = 2 e E S$

$∴ S = \frac{V}{E} = S_{0}$

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Two infinite line charges having linear charge density λ₁ and λ₂ are placed along the x and y axis respectively. The electric field at a point (x₁, y₁) in the first quadrant is inclined at an angle of 60° with the x-axis. Then which of the following relations is correct.

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Vishal Baghel

Contributor-Level 10

tan 60° = (2kλ? /y? )/ (2kλ? /x? )

(√3λ? /x? ) = (λ? /y? ); √3λ? y? = λ? x?

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

The diagram shows three concentric conducting spherical shells having radii $R, 2 R$ and $3 R$ . All three shells are given some charges. The initial potential of each shell is as mentioned in the figure. If the inner most shell is earthed then the final charge present on its outer surface would be equal to:

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Raj Pandey

Contributor-Level 9

$V_{1} = \frac{k Q_{3}}{3 R} + \frac{k Q_{2}}{2 R} + \frac{k Q_{1}}{R} = 30$

$V_{2} = \frac{k Q_{3}}{3 R} + \frac{k Q_{2}}{2 R} + \frac{k Q_{1}}{2 R} = 15$

$V_{3} = \frac{k Q_{3}}{3 R} + \frac{k Q_{2}}{3 R} + \frac{k Q_{1}}{3 R} = 10$

Eq. (1) $&$ (2) $\to \frac{k Q_{1}}{R} = 30$

Eq. (2) $&$ (3) $\to \frac{k Q_{2}}{6 R} + \frac{k Q_{1}}{6 R} = 5 \Rightarrow \frac{k Q_{2}}{R} = - \frac{k Q_{1}}{R} + 30 = 0$

By (3) $\to \frac{k Q_{3}}{R} = 30 - \frac{k Q_{2}}{R} - \frac{k Q_{1}}{R} = 0$

Potential of innermost shell (earthed) $= \frac{k Q_{1}^{'}}{R} + \frac{k Q_{2}}{2 R} + \frac{k Q_{3}}{3 R} = 0$

$\Rightarrow \frac{k Q_{1}^{'}}{R} = 0 ∴ Q_{1}^{'} = 0$

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2 months ago

Physics Electrostatic Potential and Capacitance Class 12th

Two capacitros are fully charged by a battery as shown in the figure.

If the capacitors are disconnected from battery and plate 1 is connected to plate 4 & plate 2 to plate 3, then find the potential difference (in volts) across capacitors finally.

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Vishal Baghel

Contributor-Level 10

Charge initially on capacitor 1=10µC
Charge initially on capacitor 2=20µC
Finally the potential difference across both capacitor will be equal.
Let us assume that to be V.
1V + 2V = +10
V = 10 / 3 = 3.33V

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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…………. $μ F .$