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a point charge q is located on axis of disc of radius r at distance b from the plane of disc.if one-fourth of electric flux from charge passes through disc then....... (a)r = b (b)r = 2b

a point charge q is located on axis of disc of radius r at distance b from the plane of disc.if one-fourth of electric flux from charge passes through disc then.......
(a)r = b    (b)r = 2b

Grade:12

4 Answers

Badiuddin askIITians.ismu Expert
148 Points
13 years ago

Dear Dinesh

Draw a sphere whose center is at a distance b from the disk and which cover the disk

 

7477-278_7662_untitled.JPG

we know that flux throug a segment which makes an angel  θ at the center is

Φ=Q/2ε (1-cosθ)

 

where cosθ = b/(b2+r2)1/2

given flux passes in 1/4 th

so Q/2ε (1-cosθ)  =Q/4ε

    1-cosθ = 1/2

     cosθ =1/2

  b/(b2+r2)1/2 =1/2

  r =√3 b

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Nitish Sainani
24 Points
7 years ago
(it is disk not the spherical cap how can you take a sphere)
 
Carelle
13 Points
5 years ago
A 1000w laser beam is concentrated by a Lens of a cross-sectional area of 10^-5 cm^2 . find the corresponding amplitude of the electric field
Kushagra Madhukar
askIITians Faculty 628 Points
2 years ago
Hello student
As we know total electric flux given by a point charge Q = Q/ε0            
this flux is equally distributed in a solid angle of 4π
 
Now, if a disk makes an angle θ with the axis of charge, the solid angle covered by the disk 
Ω = 2π(1- cosθ)
 
Hence electric flux through disk, ɸ =  Q/4πε0 x 2π(1- cosθ) = Q/2ε0 x (1 – cosθ)
Given, this flux is ¼ of the flux given by charge Q
Hence,
Q/2ε0 x (1 – cosθ) = Q/4ε0
or, cosθ = ½ 
or, θ = 60o
Hence, b/R = cos60o
or, b = R/2
or, R = 2b
or, b = (b2 + r2)1/2/2
or, 4b2 = b2 + r2
or, r2 = 3b2
or, r = √3 b
 
Hope it helps
Regards,
Kushagra

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