Physics Electrostatic Potential and Capacitance

Get insights from 132 questions on Physics Electrostatic Potential and Capacitance, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Electrostatic Potential and Capacitance

Follow Ask Question
132

Questions

0

Discussions

5

Active Users

0

Followers

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

For uniformly charged spherical shell,

V = k q R (  For rR )

V C = V P

VCVP= Zero 

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

(A) If c  is the velocity of light

so,   E = h ν   (Energy of photon)

(B) Velocity of photon is equal to velocity of light i.e. c.

(C)  λ = h p

p = h λ

p = h v c

(D) In photon-electron collision both total energy and total momentum are conserved.

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

V . C = M S D V S D     … (i)

given :  (N+1)VSD=NMSD

VSD= (NN+1)MSD     … (ii)

From (1) and (2)

V . C = ( MSD ) N N + 1 ( MSD )

=MSD (1NN+1)=MSDN+1

= 0 . 0 1 N + 1 = 1 1 0 0 ( N + 1 )

New answer posted

4 months ago

0 Follower 21 Views

P
Payal Gupta

Contributor-Level 10

Capacitance of each capacitor

C1=A3ε012=6Aε0

C2=A4ε0=4Aε0

Equivalent capacitance

Δv2=240Aε04Aε0=60v

vfoil=60v

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

The potential V at any point, at distance r from centre of dipole = KPcosθ r 2

At axial point where  θ=0? , V=KPr2=9*109*4*10622=9*103 V

At axial point where  θ=180? , V=KPr2=9*103 V

New answer posted

4 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

Given circuit is balanced Wheatstone bridge

C A B = 1 + 1

= 2 ? F

 

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

F = q ( Q q ) 4 π ε 0 r 2 d F d q = 1 4 π ε 0 r 2 ( Q 2 q ) d F d q = 1 2 π ε 0 r 2

Here r is fixed

For maxima or minima of force, its first derivative should be zero.

d F d q = 1 4 π ε 0 r 2 ( Q 2 q ) = 0 q = Q 2                

Since second derivative is always negative so maxima will occur at this value of q.

 

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

V = 10 ( 1 e t / R C )  

2 = 20 ( 1 e t / R C )  

1 1 0 = 1 e t / R C

e t / R C = 1 0 9

t R C = l n ( 1 0 9 ) = 0 . 1 0 5

C = t R * 0 . 1 0 5 = 1 0 6 1 0 * 0 . 1 0 5 = 0 . 9 5 μ F              

 

New answer posted

4 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Δ v = E . d x

Δ v = σ o d 2

C N e w = Q Δ v = σ A 2 o σ d = 2 o A d

C N e w = 2 C o r i g i n a l

C N e w C O r i g i n a l = 2 : 1

New answer posted

4 months ago

0 Follower 29 Views

A
alok kumar singh

Contributor-Level 10

λ = Q R * 2 π / 3

E = 2 k λ R s i n ( θ 2 ) ( i ^ )

E = 2 k R * 3 Q 2 π R s i n 6 0 ° ( i ^ )

E = 3 3 Q 8 π 2 ε 0 R 2 ( i ^ )

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 681k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.