Tags Electric Charges and Fields

Electric Charges and Fields

Get insights from 101 questions on Electric Charges and Fields, answered by students, alumni, and experts. You may also ask and answer any question you like about Electric Charges and Fields

Follow Ask Question

101

Questions

Discussions

Active Users

Followers

New answer posted

4 months 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?

0 Follower 6 Views

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

0 0

New answer posted

5 months 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.

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

$N = m g c o s 30 ° + q E s i n 30 °$

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

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

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

0 0

New answer posted

5 months 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,

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

0 Follower 5 Views

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,

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

0 0

New answer posted

5 months ago

Electric Charges and Fields Electric Charge

Why is an electric charge a scalar quantity?

0 Follower 1 View

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.

0 0

New answer posted

5 months 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.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

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

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

0 0

New answer posted

5 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 $\in_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} - m^{- 2}]$

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

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

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

0 0

New answer posted

5 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.

0 Follower 3 Views

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

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

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

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

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

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

0 0

New answer posted

5 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

0 Follower 2 Views

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.

0 0

New answer posted

5 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:

0 Follower 2 Views

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

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

0 0

New answer posted

5 months ago

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

Given below are two statements:

Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.

Statement II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r (

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

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

All the charge given to a conducting sphere resides on outer surface.

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
681k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Electric Charges and Fields

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience