Physics Electric Charge and Field

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New answer posted

2 months ago

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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

2 months ago

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A
alok kumar singh

Contributor-Level 10

  3 μ F and 3 μ F  in parallel

 

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2 months ago

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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?

New question posted

2 months ago

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New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

x=qEt²/2m, y=gt²/2. y= (mg/qE)x. Straight line.

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

x > d path is straight line
-y = 1/2 at²
x-d = V?t ⇒ t=(x-d)/V?
-y = 1/2 a((x-d)/V?)²
-y/at = (1/2at)(a²t²)
at = V?
(-y/at) = (d/2V?)
(x-d)/V? = d/V?
(-y - 1/2 a t²)/(at) = (x-d - V?t)/V?
y = (-qEd/mV?)(x/V? - d/2V?) ; y = (qEd/mV?²)(d/2 - x)

 

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Net field along AB at O must be zero.
E? cosα = E? sinα
(kQ? /x? ²) (x? /AB) = (kQ? /x? ²) (x? /AB)
Q? /Q? = x? ³/x? ³

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

E ? = σ 2 ε 0 c o s ? 60 ? ( - x ˆ ) + σ 2 ε 0 - σ 2 ε 0 s i n ? 60 ? ( y ˆ )

E ? = σ 2 ε 0 1 - 3 2 y ˆ - 1 2 x ˆ

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

Tsinθ = qE
Tcosθ = mg
tanθ = qE/mg
E = V/d_eff. V = V? - (-V? ) = V? +V?
C? =kε? A/t, C? =ε? A/ (d-t).
Capacitors are in series. Q=C_eqV. E inside dielectric = σ/ (kε? ) = Q/ (Akε? ).
E in air = σ/ε? = Q/Aε?
Let's follow the solution.
Q = (C? / (C? +C? ) (V? +V? ).
E = E_air = Q/ (Aε? )
tanθ = qQ/ (Aε? mg) = q/ (Aε? mg) * (C? / (C? +C? ) (V? +V? )
Substitute C? and C?
This is getting complicated. The solution directly uses an expression for E. E = Q/C? Aε?

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

E? (due to A and G) = 2 * (kq/l²)cos (45) = √2 kq/l² (downwards)
E? (due to B and F) = 2 * (kq/l²)cos (45) = √2 kq/l² (towards left)
E? (due to C and H) = 0
E? (due to D and E) = 0
Resultant E = √ (E? ²+E? ²) = √ (2 (kq/l²)²+2 (kq/l²)²) = 2kq/l²
(Solution in the image seems to be different.)

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