Physics Electrostatic Potential and Capacitance

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New answer posted

3 weeks ago

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R
Raj Pandey

Contributor-Level 9

3 μ F and 3 μ F  in parallel

 

New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

For a conducting sphere, the electric field E = σ/ε? and potential V = σR/ε?
When two spheres are connected by a wire, their potentials become equal: V? = V?
σ? R? /ε? = σ? R? /ε?
σ? R? = σ? R? ⇒ σ? /σ? = R? /R?

New answer posted

4 weeks ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

C = A B ? A ? B ?

Using De-Morgan Theorem

C = A B ? A ? B ?

C = B ( A + A ? ) ? = B ?

Therefore

 

New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

Time period of satellite

T = 2 π R 3 G M

= 2 π R 3 G d 4 3 π R 3

T = 3 π G d 3 π G d = T 2

New answer posted

4 weeks ago

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R
Raj Pandey

Contributor-Level 9

Electric field is always perpendicular to equipotential surface.

New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

No current through '  G  '

So potential difference across R is 2 V

i = 8 400 R = 2 8 * 400 = 100 Ω

New answer posted

4 weeks ago

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V
Vishal Baghel

Contributor-Level 10

F = (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + .

⇒ F = (qQ)/ (4πε? ) [1/1² + 1/2² + 1/4² + 1/8² + .]

⇒ F = (qQ)/ (4πε? ) [1/ (1 - 1/4)]

⇒ F = (qQ)/ (4πε? ) [4/3] = 10? * 1 * 9 * 10? * (4/3) = 12 * 10³ N

New answer posted

a month ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

The flux φ is calculated as:
φ = (2/5) * 4 * 10³ * 0.4 = 640 Nm²C? ¹

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Let the charge on C? be q µC. For the capacitor network:
q/C? = (C? * 10 - q) / C? ⇒ q/8 = (2 * 10 - q) / 2 ⇒ q = 16

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