Electric Charges and Fields

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New answer posted•a year ago
S
a year ago

The following are the topics covered in this chapter: Electric Field and Field Lines, Gauss's Law, Electric Dipole, Conductors and Insulators, and Electric Flux. 

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New answer posted•a year ago
S
a year ago

Indeed, it is an easy chapter of Class 12 Physics. In this chapter, you will learn about the foundational concepts like Gauss's law and electric fields.

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New answer posted•a year ago
S
a year ago

Flux through the 6 sides of square (i.e. cube)

? T = q i n ε 0 = 1 2 × 1 0 − 6 C ε 0   


Flux through a square

? = ? T 6 = 1 2 × 1 0 − 6 ε 0 × 6 = 2 × 1 0 − 6 ε 0

⇒ ? = 2 2 5 . 9 8 × 1 0 3 N m 2 / C ? 2 2 6 N m 2 / C     

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New answer posted•a year ago
S
a year ago

In the medical entrance test NEET, this chapter has a moderate weightage. You can expect around 2-3 questions of this chapter that contributes to the 3-5% of the total marks in the Physics section.

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New answer posted•a year ago
S
a year ago

From Guass’s Law

? E → . d A → = q i n ∈ 0            

E . 4 π r 2 = ∫ 0 r ( 4 π r 2 . d r ) ( α r ) ∈ 0            

E = ( α r 2 4 ∈ 0 )            

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

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)

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New answer posted•a year ago
S
a year ago

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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