The magnitude of the electric field at a distance r = 12 cm from the line of cha

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Electric field because of infinity long line of charge : Use gauss law to fiend it ?E . dA = q/e ; e = permitivity take a tube surface enclosing the wire, we have: E .* 2*pi*r * L = b L/ e ; b = charge per unit length E = b/(2* pi* e * r) 3 x 10^-5 = b/(2* pi* e * 0.12) => b/(2* pi* e) = 3.6 x 10^-6 .................(a) E = b/(2* pi* e * r) = b/(2* pi* e) * 1/0.04 = 3.6 x 10^-6 /0.04 = 9 x 10^-5 N/C for positive line of charge Electric field point away radially from the wire. That why we can use Gauss law here. b/(2* pi* e) = 3.6 x 10^-6 ( from a)) b = (2* pi* e) * 3.6 x 10^-6 = 2 * 3.14 * 8.85 x 10^-12 * 3.6 x 10^-6 = 2 x 10^-16 C/m Q = b * 1 m = 2 x 10^-16 C It is too small, I think you wrong write E = 3 x 10^-5 N/C on the question maybe it was 3 x 10^5 N/C. It would change the answer to E = 9 x 10^5 N/C and Q = 2 microC.

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