Electrostatics

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New answer posted•a year ago

Electrostatics Class 12th

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The potential ( $V$ ) at any axial point, at 2 m distance ( $r$ ) from the centre of the dipole of dipole moment vector $\dot{P}$ of magnitude, $4 \times 1 0^{- 6} C m$ , is $\pm 9 \times 1 0^{3} V$ .

(Take $\frac{1}{4 π ϵ_{0}} = 9 \times 1 0^{9}$ SI units)

Reason R : $V = \pm \frac{2 P}{4 π ?_{0} r^{2}}$ , where $r$ is the distance of any axial point, situated at 2 m from the centre of the dipole.

In the light of the above statements, choose the correct answer from the options given below:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

The potential $V$ at any point, at distance $r$ from centre of dipole $= \frac{K P cos θ}{r^{2}}$

At axial point where $θ = 0^{?}, V = \frac{K P}{r^{2}} = \frac{9 \times 1 0^{9} \times 4 \times 1 0^{- 6}}{2^{2}} = 9 \times 1 0^{3} V$

At axial point where $θ = 18 0^{?}, V = \frac{- K P}{r^{2}} = - 9 \times 1 0^{3} V$

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New answer posted•a year ago

Electrostatics Class 12th

In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is :

41 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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New answer posted•a year ago

Electrostatics Class 12th

A certain charge Q is divided into parts q and (Q – q). How should the charges Q and q be divided so that q and (Q – q) placed at a certain distance apart experience maximum electrostatic repulsion?

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$F = \frac{q (Q - q)}{4 π ε_{0} r^{2}} \Rightarrow \frac{d F}{d q} = \frac{1}{4 π ε_{0} r^{2}} (Q - 2 q) \Rightarrow \frac{d F}{d q} = \frac{1}{2 π ε_{0} r^{2}}$

Here r is fixed

For maxima or minima of force, its first derivative should be zero.

$\frac{d F}{d q} = \frac{1}{4 π ε_{0} r^{2}} (Q - 2 q) = 0 \Rightarrow q = \frac{Q}{2}$

Since second derivative is always negative so maxima will occur at this value of q.

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New answer posted•a year ago

Electrostatics Class 12th

Figure shows a rod AB, which is bent in a 120° circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric field $\vec{E}$ at the centre of curvature O?

111 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$λ = \frac{- Q}{R \times 2 π / 3}$

$\vec{E} = \frac{2 k λ}{R} s i n (\frac{θ}{2}) (- \hat{i})$

$\Rightarrow \vec{E} = \frac{2 k}{R} \times \frac{- 3 Q}{2 π R} s i n 60 ° (- \hat{i})$

$\Rightarrow \vec{E} = \frac{3 \sqrt[]{3} Q}{8 π^{2} ε_{0} R^{2}} (\hat{i})$

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New answer posted•a year ago

Electrostatics 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:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$Q_{1} = C_{1} V = 2 V μ C$

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

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

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New answer posted•a year ago

Electrostatics 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.

15 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Volume of 27 identical drops = volume of a bigger drop

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

R³ = 27r³

R = 3r

Given potential of a small drop = 22v

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

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New answer posted•a year ago

Electrostatics 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:

42 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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

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