Electrostatics

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

Potential of centre, = V =
Vc = K (Σq)/R
Vc = K (0)/R = 0
Electric field at centre E_B = E_B = ΣE
Let E be electric field produced by each

charge at the centre, then resultant electric field will be
Ec = 0, since equal electric field vectors are acting at equal angle so their resultant is equal to zero.

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

6

Sol. Common potential after connection.

V common   = C 1 V 1 + C 2 V 2 C 1 + C 2 = 60 × 20 + 0 120 = 10 V o l t

 

Loss of energy = 1 2 C V 2 - 1 2 ( 2 C ) × V Common   2

= 1 2 × 60 × 10 - 12 × ( 20 ) 2 - 60 × 10 - 12 × ( 10 ) 2 = 60 × 10 - 12 ( 200 - 100 ) = 6000 × 10 - 12 = 6 n J

 

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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

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

E → = 1 4 π ε 0 ( − 2 Q ) ( 3 2 a ) 2 . 1 3 ( x ^ + y ^ + z ^ ) = | − 2 Q 3 3 π ε 0 a 2 ( x ^ + y ^ + z ^ )

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New answer posted•a year ago
S
a year ago

When connected in series, equivalent capacitance,

C 1 = C * C C + C = C 2            

When connected in parallel, equivalent capacitance

C2 = C + C = 2C

C 1 C 2 = C / 2 2 C = 1 4            

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New answer posted•a year ago
S
a year ago

(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.

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New answer posted•a year ago
S
a year ago

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 (1−NN+1)=MSDN+1

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

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