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Part A To find $V$, divide the rod into infinitesimal segments of length $dx_1$. Consider one such segment located at $x = x_1$, where $-a \le x_1 \le a$. What is the potential $dV$ at $x = +L$ due to this segment? Express your answer in terms of some or all of the variables $Q$, $a$, $L$, $dx_1$, $x_1$, electric constant $\epsilon_0$, and $\pi$. dV =

          Part A
To find $V$, divide the rod into infinitesimal segments of length $dx_1$. Consider one such segment located at $x = x_1$, where $-a \le x_1 \le a$. What is the potential $dV$ at $x = +L$ due to this segment?
Express your answer in terms of some or all of the variables $Q$, $a$, $L$, $dx_1$, $x_1$, electric constant $\epsilon_0$, and $\pi$.
dV =
        
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Part A
To find V, divide the rod into infinitesimal segments of length dx1. Consider one such segment located at x = x1, where -a ≤ x1 ≤ a. What is the potential dV at x = +L due to this segment?
Express your answer in terms of some or all of the variables Q, a, L, dx1, x1, electric constant ϵ0, and π.
dV =

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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To find V, divide the rod into infinitesimal segments of length dx_(1). Consider one such segment located at x=x_(1), where -a<=x_(1)<=a. What is the potential dV at x=+L due to this segment? Express your answer in terms of some or all of the variables Q, a, L, dx_(1), x_(1), electric constant epsilon_(0), and pi.
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Transcript

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00:01 So in this question we're told that dr by dx is equal to 6x, and we're told that for a voltage of 6 .6 volts, we want the power to be 218 watts.
00:23 And the power is the voltage times the current, but the voltage is the current times the resistance.
00:31 So we can write that the current is the voltage divided by the resistance, so that the power is the voltage square divided by the resistance.
00:45 So to calculate the resistance, we want a resistance, which is the voltage square divided by the power...
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