Physics Electric Charge and Field

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New answer posted

a week ago

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A
alok kumar singh

Contributor-Level 10

From Gauss law

? = q e n c l o s e d ε 0 = 5 q ε 0

New answer posted

a week ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

For maximum value of s, initially, electron must move away from plate.

ut + 1 2 at2 = s

t = 1u = 1m/s               s = –1 m

1 * 1 – 1 2 a * 12 = – 1

-> a = 4m/s2

q E m = 4      

  q σ 2 ε 0 m = 4     

σ = 4 * 2 * 9 * 1 0 1 2 * 9 * 1 0 3 1 1 . 6 * 1 0 1 9

= 8 * 8 1 1 . 6 * 1 0 2 4      

= 4.05 * 10–22C/m2

New answer posted

2 weeks ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

2 weeks ago

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A
alok kumar singh

Contributor-Level 10

Flux through a surface ? = E ? . A ?

? = 6 i ? - 4 j ? + 7 k ? . 200 i ?

= 1200 u n i t

New answer posted

2 weeks ago

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R
Raj Pandey

Contributor-Level 9

Because E points along the tangent to the lines of force. If initial velocity is zero then due to the force, it always moves in the direction of E. Hence will always move on some lines of force

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2 weeks ago

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R
Raj Pandey

Contributor-Level 9

In Case I when both are positively charged, due to induction positive charge moves outwards on spheres, increasing effective distance between centres of charge causing magnitude of the force to decrease.

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A
alok kumar singh

Contributor-Level 10

F21 = QE1 = λ2l (2kλ1/R)

New answer posted

2 weeks ago

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A
alok kumar singh

Contributor-Level 10

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

New answer posted

2 weeks ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

  N = m g c o s 3 0 ° + q E s i n 3 0 °

a = m g s i n θ q E c o s θ μ N m = 2 . 3 0 m / s 2

S = u t + 1 2 a t 2

t = 2 l a = 1 . 3 1 s e c             

New answer posted

2 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

? 1 = 3 5 E 0 ( 0 . 2 ) N m 2 C 1 , a n d ? 2 = 4 5 E 0 ( 0 . 3 ) N m 2 C 1

? 1 ? 2 = 3 * 0 . 2 * 5 5 * 0 . 3 * 4 = 1 2

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