A charge of 5.0 C is kept at the centre of a sphere of radius 1 m. What is the flux passing through the sphere? How will this value change if the radius of the sphere is doubled?
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Given: q = 5C, r = 1 m
To find: Flux (ø)
Formulae: i. E = \(\frac{1}{4\pi \epsilon_0}\times\frac{q}{r^2}\)
ii. ø = E × A = E (4πr2)
Calculation: From formula (i),
E = 9 × 109 × \(\frac{5}{1^2}\)
= 4.5 × 1010 N/C
From formula (ii),
ø = E × 4 π r2
= 4.5 × 1010 × 4 × 3.14 × 12
ø = 5.65 × 1011 Vm
This value of flux will not change if radius of sphere is doubled. Though radius of sphere will increase, increased distance will reduce the electric field intensity. As E ∝ \(\frac{1}{r^2}\) and A × r2 net variation in total flux will not be observed.
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