Two point charges q and –q are separated by the distance 2 λ (Figure). Find the flux of the electric field strength vector across a circle of radius R.
Text Solution
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The sign of φ φ depends on how the direction of the normal to the circle is chosen.
Sol.
Electric field at distance y on the circle due to both charges is
E = 2 × 
× cosθ= 
× 
E = 
(Along the dotted line)
flax through the width dy of circle d φ φ = E (2πy. dy) (Angle = 0º)
= 2π(2 kq λ ) 
Let λ 2 + y 2 = x
∴ 2ydy = dx
= 
= 
. 
= – 
= 
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