(Be dear and concise. Typing work is strongly suggested. Will thumps up dependin

Question

(Be dear and concise. Typing work is strongly suggested. Will thumps up depending on the quality of work done.) When a potential difference of 12 V is applied to a wire 7.6 m long and 0.42 mm in diameter the result is an electric current of 1.1 A. What is the resistivity of the wire? To operate a given flash lamp requires a charge of 20pC. What capacitance is needed to store this much charge in a capacitor with a potential difference between its plates of 3.1 V? How far must the point charges q1 = +7.22 mu C and q_2 = -36.6 mu C be separated For the electric potential energy of the system to be -169 J?

Explanation / Answer

Here ,

1a)

V = 12 V

L = 7.6 m

d = 0.42 mm

current , I = 1.1 A

for the resistivity is p

as R = V/I = p * L/Area

12/1.1 = p * 7.6/(pi * (0.00042/2)^2)

solving for p

p = 1.99 *10^-7 Ohm .m

the resistivity of wire is 1.99 *10^-7 Ohm .m

1b)

as Q = C * V

20 uC = C * 3.1

C = 6.45 uF

the capacitance needed is 6.45 uF

1c)

let the distance between the charges is d

Potential energy = k * q1 * q2/d

-169 = 7.22 *10^-6 * (-36.6 *10^-6) * 9*10^9/d

solving for d

d = 0.0141 m

the distance between the charges is 0.0141 m

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