Electric Charges and Fields

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New answer posted•a year 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?

15 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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New answer posted•a year ago

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.

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

13 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$N = m g c o s 30 ° + q E s i n 30 °$

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

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

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

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New answer posted•a year 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, 

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

19 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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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New answer posted•a year ago

Electric Charges and Fields Electric Charge

Why is an electric charge a scalar quantity?

1 Views|Posted a year ago

Asked by Satyendra

Answered by

Pallavi Arora | Beginner-Level 5

a year ago

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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New answer posted•a year ago

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.

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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New answer posted•a year 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 $\in_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} - m^{- 2}]$

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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New answer posted•a year 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.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

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

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

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New answer posted•a year 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

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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•a year 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:

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Since, volume remain constant

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

R³ = r³ × 64

R = 4r

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