Physics Electric Charge and Field

Get insights from 91 questions on Physics Electric Charge and Field, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Electric Charge and Field

Follow Ask Question
91

Questions

0

Discussions

9

Active Users

1

Followers

New answer posted

3 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

4 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

4 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

New answer posted

4 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

According to Lorentz's Force, we can write

F=FElectric+FMagnetic=qE+q (v*B), so

Statement I is correct but Statement II is incorrect

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

First resonating length = λ = V f = 3 4 0 3 4 0 = 1 m

Second resonating length = 3 λ 4 = 7 5 c m

Third resonating length = 5 λ 4 = 1 2 5 c m

Height of water required = 125 – 75 = 50cm

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Q = C1v                                                                                                          

...more

New answer posted

4 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Please find the solution below:

 

New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

E1 (done to q1) = k, q1 (d/2)2= [9*109*8*106]d2*4

E2 (done to q2) = kq2 (d/2)2= [9*109*8*106]d2*4

Enet = E1 + E2

=2 [4*9*109*8*106d2]

Given Enet = 6.4 * 104

2* [4*9*109*8*106d2]=6.4*104

d2 = 9 * 10-6 + 6

d = 3m

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

E1 (done to q1) = k, q1 (d/2)2= [9*109*8*10? 6]d2*4

E2 (done to q2) = k? q2 (d/2)2= [9*109*8*10? 6]d2*4

Enet = E1 + E2

=2 [4*9*109*8*10? 6d2]

Given Enet = 6.4 * 104

? 2* [4*9*109*8*10? 6d2]=6.4*104

d2 = 9 * 10-6 + 6

d = 3m

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

F1=9*109*5*106*0.3*1069*104

= 15 N

F2=9*109*5*106*0.16*1069*104

= 8N

Fnet= (15)2+ (8)2

= 17 N

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

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 679k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.