If the linear charge density of a cylinder is 4 μ cm-1 then electric field intensity at point 3.6 cm from axis is

Question

If the linear charge density of a cylinder is 4 μ cm-1 then electric field intensity at point 3.6 cm from axis is (A)  4 × 105 NC-1  (B)  2 × 106 NC-1  (C)  8 × 107 NC-1  (D)  12 × 107 NC-1

Tardigrade Admin · Accepted Answer

For a charged cylinder E =(λ/2 π ε0 r) =(4 × 10-6 × 18 × 109/3.6 × 10-2) =2 × 106 NC -1

Q. If the linear charge density of a cylinder is $4 μ c m^{- 1}$ then electric field intensity at point $3.6 c m$ from axis is

$4 \times 1 0^{5} N C^{- 1}$

$2 \times 1 0^{6} N C^{- 1}$

$8 \times 1 0^{7} N C^{- 1}$

$12 \times 1 0^{7} N C^{- 1}$

For a charged cylinder
$E = \frac{λ}{2 π ε _{0} r}$
$= \frac{4 \times 1 0 ^{- 6} \times 18 \times 1 0 ^{9}}{3.6 \times 1 0 ^{- 2}}$
$= 2 \times 1 0^{6} N C^{- 1}$

Q. If the linear charge density of a cylinder is 4μcm−1 then electric field intensity at point 3.6cm from axis is

4×105NC−1

2×106NC−1

8×107NC−1

12×107NC−1