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Electric Charges and Fields

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Electric Charges and Fields Class 12th

What will be the magnitude of electric field at point O as shown in figure? Each side of the figure is l and perpendicular to each other?

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Vishal Baghel

Contributor-Level 10

E? (due to A and G) = 2 * (kq/l²)cos (45) = √2 kq/l² (downwards)
E? (due to B and F) = 2 * (kq/l²)cos (45) = √2 kq/l² (towards left)
E? (due to C and H) = 0
E? (due to D and E) = 0
Resultant E = √ (E? ²+E? ²) = √ (2 (kq/l²)²+2 (kq/l²)²) = 2kq/l²
(Solution in the image seems to be different.)

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New answer posted

a month ago

Electric Charges and Fields Class 12th

Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical?

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Vishal Baghel

Contributor-Level 10

T sin θ = (1/4πε? ) * q²/ (2lsinθ)²
T cos θ = mg
∴ tan θ = q² / (4πε? mg * 4l²sin²θ)
[tan θ ≈ θ, for small angle]
So, θ³ = q² / (16πε? mgl²)
θ = ( q² / (16πε? mgl²) )¹/³
Also separation = 2l sin θ ≈ 2lθ
= 2l ( q² / (16πε? mgl²) )¹/³
= ( 8q²l³ / (16πε? mgl²) )¹/³
= ( q²l / (2πε? mg) )¹/³

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Electric Charges and Fields Class 12th

An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field $200 \frac{N}{C}$ as shown in the figure. A body of mass 1kg and charge 5 mC is allowed to slide down from rest at a height of 1m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom.

$[g = 9.8 m / s^{2}; s i n 30 ° = \frac{1}{2}; c o s 30 ° = \frac{\sqrt[]{3}}{2}]$

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Vishal Baghel

Contributor-Level 10

$N = m g c o s 30 ° + q E s i n 30 °$

$a = \frac{m g s i n θ - q E c o s θ - μ N}{m} = 2.30 m / s^{2}$

$S = u t + \frac{1}{2} a t^{2}$

$\Rightarrow t = \sqrt[]{\frac{2 l}{a}} = 1.31 s e c$

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a month ago

Electric Charges and Fields Class 12th

In an octagon ABCDEFGH of equal side, what is the sum of $\vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F} + \vec{A G} + \vec{A H},$ if,

$\vec{A O} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$

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Vishal Baghel

Contributor-Level 10

$\vec{A B} + \vec{A H} = \vec{A O}$

$\vec{A C} + \vec{A G} = 2 \vec{A O}$

$\vec{A D} + \vec{A F} = 3 \vec{A O}$

$\vec{A E} = 2 \vec{A O}$

Adding All,

$\vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F} + \vec{A G} + \vec{A H} = 8 \vec{A O} = 16 \hat{i} + 24 \hat{j} - 32 \hat{k}$

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a month ago

Electric Charges and Fields Electric Charge

Why is an electric charge a scalar quantity?

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Pallavi Arora

Beginner-Level 5

When two or more individual charges are present in a system, the total charge will be an algebraic sum of all individual charges and not the vector sum. Therefore, an electric charge is considered as a scalar quantity.

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Electric Charges and Fields Class 12th

An electric dipole is placed on x – axis in proximity to a line charged of linear density 3.0 * 10^-⁶ C/m. Line charge is placed on z – axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4N is exerted on the dipole, find out the amount of positive charge of the dipole.

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Vishal Baghel

Contributor-Level 10

$F = [\frac{2 k λ}{r_{1}} - \frac{2 k λ}{r_{2}}] q = 2 k λ q [\frac{1}{r_{1}} - \frac{1}{r_{2}}]$

$2 * 9 * 1 0^{9} * 3 * 1 0^{- 6} [\frac{1000}{10} - \frac{1000}{12}] q$

$\begin{array}{l} \Rightarrow 4 = 9 * 1 0^{5} q \\ \Rightarrow q = 4.44 μ C \end{array}$

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New answer posted

2 months ago

Electric Charges and Fields Class 12th

The volume charge density of a sphere of radius 6m is 2 $μ C c m^{- 3} .$ The number of lines of force per unit surface area coming out from the surface of the sphere is _________ * 10¹⁰ NC^-1.

[Given : Permittivity of vacuum $\in_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} - m^{- 2}]$

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Vishal Baghel

Contributor-Level 10

No. of electric field line per unit area = electric field

$E = \frac{ρ r}{3 \in_{0}} f o r r = R$

$E = 45 * 1 0^{10} N / C$

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New answer posted

2 months ago

Electric Charges and Fields physics ncert exemplar solution class 12th chapter one

A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.

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Payal Gupta

Contributor-Level 10

Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i0 $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

= $= - L i_{0} \frac{- 1}{2} e^{- t / τ}$

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

$e_{a v g} = \frac{\int_{0}^{200 μ s} e d t}{\int_{0}^{200 μ s} d t} = \frac{\int_{0}^{200 μ s} \frac{L i_{0}}{τ} e^{- t / τ} d t}{\int_{0}^{200 μ s} d t}$

$= \frac{L i_{0}}{200 * 1 0^{- 6} s e c} = \frac{20 * 1 0^{- 3} * 2}{200 * 1 0^{- 6}} = 400$

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New answer posted

2 months ago

Electric Charges and Fields physics ncert exemplar solution class 12th chapter one

In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size $\frac{λ}{100},$ where $λ$ is the wavelength of the wave in free space. The phenomenon, which happens there will be

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Payal Gupta

Contributor-Level 10

If size of object is very small as compare to wave length of EM wave in free space then, scattering will happen.

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New answer posted

2 months ago

Electric Charges and Fields physics ncert exemplar solutions class 12th chapter two

Sixty four conducting drops each of radius 0.02m and each carrying a charge of $5 μ C$ are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:

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Vishal Baghel

Contributor-Level 10

Since, volume remain constant

$64 * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$

R³ = r³ * 64

R = 4r

$\frac{σ_{b i g g e r}}{σ_{s m a l l d r o p}} = \frac{5 * 64 * 4 π^{2}}{4 π R^{2} * 5} = 64 {(\frac{r}{R})}^{2} = 64 {(\frac{r}{4 r})}^{2} = \frac{64}{16} = 4 : 1$

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