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Electric Charge and Field

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a month ago

Electric Charge and Field Class 12th

A solid sphere of radius 4a with centre at origin. Two charge, –2q at (–5a, 0) and 5q at (3a, 0) is placed. Flux through sphere is $\frac{x q}{ε_{0}}$ . Find x

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alok kumar singh

Contributor-Level 10

From Gauss law

$? = \frac{q_{e n c l o s e d}}{ε_{0}} = \frac{5 q}{ε_{0}}$

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a month ago

Electric Charge and Field Class 12th

An electron is moving with speed of 1 m/s at distance of 1 m from a large sheet of charge with density sC/m². Find maximum value ofs such that electron hit the sheet after 1 sec.

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alok kumar singh

Contributor-Level 10

For maximum value of s, initially, electron must move away from plate.

ut + $\frac{1}{2}$ at² = s

t = 1u = 1m/s s = –1 m

1 * 1 – $\frac{1}{2}$ a * 1² = – 1

-> a = 4m/s²

$\frac{q E}{m} = 4$

$\frac{q σ}{2 ε_{0} m} = 4$

$σ = \frac{4 * 2 * 9 * 1 0^{- 12} * 9 * 1 0^{- 31}}{1.6 * 1 0^{- 19}}$

$= \frac{8 * 81}{1.6} * 1 0^{- 24}$

= 4.05 * 10^–22C/m²

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a month ago

Electric Charge and Field Class 12th

The electric field in a region is given by $\vec{E} = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}) \frac{N}{C} .$ The ratio of flux of reported field through the rectangular surface of are 0.2m² (parallel to y – z plane) to the of the surface of are 0.3m² (parallel to x – z plane) is a : b, where a = ……………

[Here $\hat{i}, \hat{j} a n d \hat{k}$ are the unit vectors along x, y and z – axes respectively]

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Vishal Baghel

Contributor-Level 10

$?_{1} = \frac{3}{5} E_{0} (0.2) N m^{2} C^{- 1}, a n d ?_{2} = \frac{4}{5} E_{0} (0.3) N m^{2} C^{- 1}$

$\Rightarrow \frac{?_{1}}{?_{2}} = \frac{3 * 0.2 * 5}{5 * 0.3 * 4} = \frac{1}{2}$

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a month ago

Electric Charge and Field NEET

What is the weightage of Electric Charges and Fields in CBSE Board exam?

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Pallavi Pathak

Contributor-Level 10

Class 12 physics chapter 1 Electric Charges and Fields is an important chapter for the CBSE Board exam. In unit 1 of the CBSE Board exam of class 12 Physics, there are questions from three chapters including - Electric Charges and Fields, Electrostatics, and Electrostatic Potential and Capacitance. This unit carries a total 16 marks.

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New answer posted

a month ago

Electric Charge and Field 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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Vishal Baghel

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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a month ago

Electric Charge and Field Class 12th

A charge q of mass m enters a region of electric field E = -E₀ĵ with a velocity v₀î + v₀ĵ at origin. Find the x coordinate where the charge hits the x axis again. (Neglect effect of gravity)

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Vishal Baghel

Contributor-Level 10

In y axis u? = v? a? = -E? q / m
s? = 0, u? t + (1/2)a? t² = 0 ⇒ t = 2u? /a? = 2v? m/E? q
x coordinate at that time = v? * t = (2v? m/0) * v? = (2v? ²m)/E? q

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New answer posted

a month ago

Electric Charge and Field Class 12th

A short dipole of dipole moment P is placed along x – axis at a distance d from a long wire having uniform linear charge density +λ (wire is along y – axis) as shown. Find minimum work required to align this dipole along y axis (so that dipole moment P is along +y direction)

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Vishal Baghel

Contributor-Level 10

W = Uf - Ui
= 0 - (-P (λ/2πε? d)
= Pλ/2πε? d

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New answer posted

2 months ago

Electric Charge and Field Class 12th

Charge Q? and Q? are at point A and B of a right angle triangle OAB (see figure). The resultant electric field at point O is perpendicular to the hypotenuse, then Q?/Q? is proportional to:

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Vishal Baghel

Contributor-Level 10

Net field along AB at O must be zero.
E? cosα = E? sinα
(kQ? /x? ²) (x? /AB) = (kQ? /x? ²) (x? /AB)
Q? /Q? = x? ³/x? ³

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2 months ago

Electric Charge and Field Class 12th

Two infinite planes each with uniform surface charge density $+ σ$ are kept in such a way that the angle between them is $30^{?}$ $^{}$ . The electric field in the region shown between them is given by:

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Vishal Baghel

Contributor-Level 10

$\overset{?}{E} = \frac{σ}{2 ε_{0}} c o s ? 60^{?} (- \hat{x}) + [\frac{σ}{2 ε_{0}} - \frac{σ}{2 ε_{0}} s i n ? 60^{?}] (\hat{y})$

$\overset{?}{E} = \frac{σ}{2 ε_{0}} [(1 - \frac{\sqrt[]{3}}{2}) \hat{y} - \frac{1}{2} \hat{x}]$

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New answer posted

2 months ago

Electric Charge and Field Class 12th

Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is a = $\frac{\sqrt[]{3}}{2} L .$

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Payal Gupta

Contributor-Level 10

$E = \frac{λ}{4 π ε_{0} a} [s i n α + s i n α] = \frac{λ s i n α}{2 π ε_{0} a}$

$\Rightarrow E = \frac{Q}{L 2 π ε_{0} * (\frac{\sqrt[]{3} L}{2})} * \frac{1}{2} = \frac{Q}{2 \sqrt[]{3} π ε_{0} L^{2}}$

$t a n α = \frac{\frac{L}{2}}{\sqrt[]{3} L / 2} \Rightarrow α = \frac{π}{6} \Rightarrow s i n α = \frac{1}{2}$

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