Physics Electric Charge and Field

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New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

F = [ 2 k λ r 1 2 k λ r 2 ] q = 2 k λ q [ 1 r 1 1 r 2 ]

2 * 9 * 1 0 9 * 3 * 1 0 6 [ 1 0 0 0 1 0 1 0 0 0 1 2 ] q

4 = 9 * 1 0 5 q q = 4 . 4 4 μ C

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

In given circuit inductor behave as a simple wire so resultant circuit will be

Ref = 2 + 1 = 3 Ω  

V = IR


l = 3 0 3 = 1 0 A  

 

New answer posted

2 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m      So

E = 1 5 0 * ( 0 . 5 ) 2 = 1 5 0 4

f l u x f l o w i n g ? = E A = 1 5 0 4 * ( 0 . 5 ) 2              

= 1 5 0 1 6             

Gausses law ? =qε0  

  1 5 0 1 6 = q ε 0            

= 8 . 3 * 1 0 1 1 C              

New answer posted

2 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

When electric field is parallel then it would provide zero flux.

New answer posted

2 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Viscous force = Weight

F V = ρ * 4 3 π r 3 * g = 3 . 9 * 1 0 N 1 0

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

dA = 2r dr

d q = σ * 2 π r d r

E = 0 R k d q z ( z 2 + r 2 ) 3 2

E = σ 2 ε 0 [ 1 z R 2 + z 2 ]

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

(Independent of distance)

E 1 = E 2 = σ 2 0

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

x=2ε0ρ

E2πxl=ρπx2lε0

Ex=ρx22xε0=ρx2ε0

=ρ2ε0*2ε0ρ=1v/m

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

P 1 = 1 . 2 * 1 0 3 0 C . m P 2 = 2 . 4 * 1 0 3 0 C . m

E 1 = 5 * 1 0 4 N C 1 E 2 = 1 5 * 1 0 4 N C 1

So, τ 1 τ 2 = P 1 E 1 s i n 9 0 ° P 2 E 2 s i n 9 0 ° = ( 1 . 2 * 1 0 3 0 ) * ( 5 * 1 0 4 ) ( 2 . 4 * 1 0 3 0 ) * ( 1 5 * 1 0 4 )

= 1 2 * 1 3 = 1 6 x = 6

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

According to relation between field and potential, we can write

E=dVdx (i^)=d (3x2)dx (i^)=6x (i^)

E (1, 0, 3)=6 (i^)N/C

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