Physics Electric Charge and Field
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New answer posted
4 months agoContributor-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 answer posted
4 months agoContributor-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
4 months agoContributor-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
4 months agoContributor-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.)
New answer posted
4 months agoContributor-Level 9
F_net = -kq²/ (l+x)² + kq²/ (l-x)²
= kq² [ (l+x)² - (l-x)² / (l²-x²)² ]
= kq² [ 4lx / (l²-x²)² ]
For x << l,
ma ≈ -4kq²x / l³
a = - (4kq²/ml³)x
ω² = 4kq²/ml³
ω = √ (4kq²/ml³)
= √ (4 * 9*10? * 10 / (1*10? * 1)
= 6 * 10? rad/s
= 6000 * 10? rad/s

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