Electrostatics

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New answer posted

3 months ago

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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

3 months ago

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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

3 months ago

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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

3 months ago

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A
alok kumar singh

Contributor-Level 10

Given circuit is balanced Wheatstone bridge

C A B = 1 + 1

= 2 ? F

 

New answer posted

3 months ago

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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

3 months ago

0 Follower 12 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 ^ )

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Q 1 = C 1 V = 2 V μ C

Q 2 = Q 3 = C 2 C 3 C 2 + C 3 V = 6 * 1 2 6 + 1 2 V = 4 V μ C

Q 1 : Q 2 : Q 3 = 1 : 2 : 2

New answer posted

3 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Volume of 27 identical drops = volume of a bigger drop

2 7 * 4 3 π r 3 = 4 3 π R 3              

R3 = 27r3

R = 3r

Given potential of a small drop = 22v

V b i g g e r = k ( 2 7 q ) R = 2 7 k q 3 r = 9 k q r = 9 * 2 2 = 1 9 8 v o l t

               

New answer posted

3 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

F R = { k Q q 0 ( a x ) 2 k Q q 0 ( a + x ) 2 }

= k Q q 0 { a 2 + x 2 + 2 a x a 2 x 2 + 2 a x ( a 2 x 2 ) }

F R = k Q q 0 ( 4 a x ) ( a 2 x 2 ) 2

T = 2 π a 3 m 4 k Q q 0 = 4 π 2 a 3 m * 4 π ε 0 4 Q q 0 = 4 π 3 ε 0 m a 3 Q q 0

New answer posted

4 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

μmg=kq1q2L2

0.25*0.01*10=9*109*2*107*2*107L2

=9*22*105+425*10

L = 12 cm

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