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Electrostatic Potential

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a month ago

Electrostatic Potential Class 12th

A particle of charge q and mass m is subjected to an electric field E = E₀(1 − ax²) in the x-direction, where a and E₀ are constants. Initially the particle was at rest at x = 0. Other then the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

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Vishal Baghel

Contributor-Level 10

E = E? (1 − ax²)

F = qE?
acceleration = F/m = (qE? /m) (1 - ax²) = v (dv/dx)
(qE? /m) ∫? (1 - ax²)dx = ∫? vdv; (qE? /M) (x - ax³/3) = 0
x (1 - ax²/3) = 0; x = 0 & x = √ (3/a)

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Electrostatic Potential Class 12th

A capacitor of capacitance C = 1µF is suddenly connected to a battery of 100 volt through a resistance R = 100Ω. The time taken for the capacitor to be charged to get 50 V is: [Take ln 2 = 0.69]

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Vishal Baghel

Contributor-Level 10

V = E (1 - e? /τ)
Where τ = RC = 100 * 10? = 10? sec
50V = 100V (1 - e? /10? )
1/2 = 1 - e? ¹?
1/2 = e? ¹?
-ln2 = -10? t
t = ln2/10?
t = 0.693 * 10? sec

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2 months ago

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

A two point charges $4 q$ and $- q$ are fixed on the $x$ -axis at $x = - \frac{d}{2}$ and $x = \frac{d}{2}$ , respectively. If a third point charge ' $q$ ' is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will:

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alok kumar singh

Contributor-Level 10

$U_{initial} = \frac{k (4 q) (q)}{(d / 2)} + \frac{k (q) (- q)}{(d / 2)}$

$\begin{array}{r} \Rightarrow \frac{6 k q^{2}}{d} \\ \Rightarrow U_{final} = \frac{4 (4 q) (q)}{(\frac{3 d}{2})} + \frac{k (q) (- q)}{(d / 2)} \\ \Rightarrow \frac{2}{3} \frac{k q^{2}}{d} \\ \Rightarrow Δ U = (\frac{2}{3} - 6) \frac{k q^{2}}{d} \Rightarrow \frac{- 16}{3} \frac{{k q}^{2}}{d} \end{array}$

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New answer posted

2 months ago

Electrostatic Potential physics ncert exemplar solutions class 12th chapter two

A air bubble of radius 1 cm in water has an upward acceleration 9.8 cm^-2 . The density of water is 1 gmcm^-3 and water offers negligible drag force on the bubble. The mass of the bubble is: (g=980 cm/s²)

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alok kumar singh

Contributor-Level 10

$ρ v g - m g = m a$

$\Rightarrow \frac{ρ v g}{m} = g + a$

$\Rightarrow m = \frac{ρ v g}{g + a}$

$\Rightarrow \frac{10^{3} (\frac{4}{3} π * 10^{- 6}) (9.8)}{9.898}$

$\Rightarrow 4.15 g m$

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