low-flow pipe..Physics question…please help!!!?

Question

An aluminum "12-gauge" wire has a diameter (d = 0.205 \, 	ext{cm}). The resistivity (ho) of aluminum is (2.75 	imes 10^{-8} \, \Omega \cdot 	ext{m}). The electric field in the wire varies with time according to the equation (E(t) = 0.0004t^2 - 0.0001t + 0.0004 \, 	ext{N/C}), where (t) is measured in seconds.

Find the current (I) through the conductor at (t = 5.0 \, 	ext{s}). (I have already solved this part and found the answer to be (1.2 \, 	ext{A}).)

Find the total charge (Q) passing through a cross-section of the conductor between (t = 0 \, 	ext{s}) and (t = 5.0 \, 	ext{s}). (I am struggling with this part and cannot seem to get the correct answer.)

Anonymous · Accepted Answer

You have to find an expression for the current as a function of time. You Know that current is a measure of the rate at which charge passes through a wire’s area, basically I = dQ/dt, taking this you can develop an expression for Q = ∫ I(t)*dt. From doing the first part, you must have gotten the an expression where I = EA/p? Assuming you did, you can plug that in for I, and solve taking the antiderivative.
Q = ∫ I(t)*dt = ∫ E*A/p, where A/p is a constant allowing you to pull it ouside the integral
Q = A/p ∫E, once u get to this point, you take the antiderivative of E
Q = ((pi*r^2)/p) (E antiderivative)
Q = ((3.14*0.001025^2)/(2.75 X 10^-8)) * (0.0004t^3/3 – 0.0001t^2/2 +0.0004t)
Q = 120 * [(0.0004t^3/3 – 0.0001t^2/2 +0.0004t)] where t=5
Q = 2.09 C

Anonymous

low-flow pipe..Physics question…please help!!!?

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