Answer:
given,
radius of rod = 5 cm
charge per unit length (λ) = 30.0 n C/m = 30 × 10⁻⁹ C/m
electric field duel to long straight wire =[tex]E =\dfrac{\lambda}{2\pi \varepsilon\ r}[/tex]
a) distance = 3 + 5 = 8 cm
[tex]E =\dfrac{30\times 10^{-9}}{2\pi \varepsilon\ 0.08}[/tex]
[tex]\dfrac{1}{2\pi \varepsilon}[/tex] = 18 × 10⁹
[tex]E =\dfrac{30\times 10^{-9}\times 18 \times 10^{9}}{ 0.08}[/tex]
E = 6750 N/C
b) distance = 10 + 5 = 15 cm
[tex]E =\dfrac{30\times 10^{-9}}{2\pi \varepsilon\ 0.15}[/tex]
[tex]\dfrac{1}{2\pi \varepsilon}[/tex]= 18 × 10⁹
[tex]E =\dfrac{30\times 10^{-9}\times 18 \times 10^{9}}{ 0.15}[/tex]
E = 3600 N/C
c) distance = 100 + 5 = 105 cm
[tex]E =\dfrac{30\times 10^{-9}}{2\pi \varepsilon\ 1.05}[/tex]
[tex]\dfrac{1}{2\pi \varepsilon}[/tex] = 18 × 10⁹
[tex]E =\dfrac{30\times 10^{-9}\times 18 \times 10^{9}}{ 1.05}[/tex]
E =514.28 N/C
