A long conducting cylinder of length and radius a (a << l) carrying a total charge q is surrounded by a conducting shell of length l and radius 2a having a total charge -2q Assume the reference of electric potential at r = 2a. Mar the CORRECT option(s): -

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The electric field at $r = \frac{3 a}{2} is \frac{4 kq}{3 aℓ} .$

The electric potential at $r = \frac{5 a}{2} is - \frac{2 kq}{ℓ} \ln (\frac{5}{4})$

The electric field at $r = \frac{5 a}{2} is - \frac{4 kq}{5 al} .$

The electric potential at $r = \frac{3 a}{2} is \frac{2 ka}{ℓ} \ln (\frac{4}{3})$

answer is A, B, C.

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Detailed Solution

$E = 0 for r < a \begin{array}{l} = \frac{2 k (\frac{q}{ℓ})}{r} for a < r < 2 a \\ = \frac{- 2 k (\frac{q}{ℓ})}{r} for r > 2 a \end{array}$
If reference is at r = 2a, then for $r = \frac{3 a}{2}, E = \frac{2 k (\frac{q}{ℓ})}{(\frac{3 a}{2})} = \frac{4 ka}{3 aℓ}$
$for r = \frac{5 a}{2}, E = \frac{- 2 k (\frac{q}{ℓ})}{(\frac{5 a}{2})} = \frac{- 4 kq}{5 aℓ} for (\frac{3 a}{2}), V_{1} - 0 = - \int_{2 a}^{\frac{3 a}{2}} \frac{2 k (\frac{q}{ℓ})}{r} dr V_{1} = 2 k {(\frac{q}{ℓ})}^{\frac{3 a}{2}} \int_{2 a}^{dr} \frac{2 k (q)}{r} = \frac{4}{ℓ} (\frac{4}{3}) For (\frac{5 a}{2}), V_{1} - 0 = 2 k (\frac{q}{ℓ}) \ln (\frac{5}{4})$