Physics Electric Charge and Field

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New answer posted

3 weeks ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

According to question, we can write

F C A = F C B = K q Q x 2 + ( d 2 ) 2 F = 2 F C A c o s θ = 2 K q Q x [ x 2 + ( d 2 ) 2 ] 3 2                         

For maxima of force  d F d x = 0 , s o  

x = d 2 2

 

New answer posted

3 weeks ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

mg In equilibrium, Fe=T sin θ

mg=T cos θ

 tan θ=Femg=q24π0x2*mg

also  tan θs?=x/2l

Hence, x 2 l = q 2 4π0x2*mg

x 3 = 2 q 2 p 4πε0mg

x=(q2l2π0mg)1/3

Therefore xl1/3

New answer posted

3 weeks ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

In y axis u? = v? a? = -E? q / m
s? = 0, u? t + (1/2)a? t² = 0 ⇒ t = 2u? /a? = 2v? m/E? q
x coordinate at that time = v? * t = (2v? m/0) * v? = (2v? ²m)/E? q

New answer posted

4 weeks ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

(Independent of distance)

E 1 = E 2 = σ 2 0

New answer posted

4 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

W = Uf - Ui
= 0 - (-P (λ/2πε? d)
= Pλ/2πε? d

New answer posted

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

? E ? d s ? = 0 ? net   = ? in   - ? out   = 0

? in   = ? out  

New answer posted

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Polar molecules have centres of positive and negative charges separated by some distance, so they have permanent dipole moment

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

i m a x = E m a x R = N B A ω R

i m a x = 1000 * 2 * 10 - 5 * π 10 2 * 2 12.56

i m a x = 1 A

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

  3 μ F and 3 μ F  in parallel

 

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

qE = mg ⇒ neE = (4/3)πr³ρg
⇒ n = (4πr³ρg) / (3eE) = (4 * 3.14 * (2 * 10? ³ )³ * 3 * 10³ * 9.81) / (3 * 1.6 * 10? ¹? * 3.55 * 10? ) = 1.73 * 10¹?

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