AP Physics C- E&M- 8.5 Electric Flux - Exam Style questions - FRQs- New Syllabus
Question
• Draw a bar to represent the relative absolute value of the work \(W_{ED}\) done by the external force on the charge to move the test charge from Position \(E\) to Position \(D\).
• Draw a bar to represent the relative absolute value of the work \(W_{DA}\) done by the external force on the charge to move the test charge from Position \(D\) to Position \(A\).
• The height of each bar should be proportional to the value of \(W_{CE}\). If \(W_{ED}=0\) and/or \(W_{DA}=0\), write a \(0\) in the corresponding column, as appropriate.
_____ \(-x\) _____ \(-y\) _____ Directly toward the sphere
ii. On Figure \(4\), sketch a graph of the \(x\)-component \(E_x\) of the electric field from the rod as a function of \(x\) in the region \(4L<x<12L\).
Most-appropriate topic codes (AP Physics C: Electricity and Magnetism):
• Topic \(8.5\) — Electric Flux (Part \( \mathrm{a} \))
• Topic \(8.6\) — Gauss’s Law (Part \( \mathrm{a} \))
• Topic \(9.2\) — Electric Potential (Parts \( \mathrm{b} \), \( \mathrm{c} \), \( \mathrm{d} \))
▶️ Answer/Explanation
(a)
Use Gauss’s law:
\(\Phi_E=\dfrac{Q_{\text{enc}}}{\varepsilon_0}\)
The \(-20.0\,\text{V}\) equipotential line encloses the charged sphere. The charge enclosed is
\(Q_{\text{enc}}=-2.0\times10^{-9}\,\text{C}\)
The problem asks for the absolute value of the flux:
\( |\Phi_E| = \left|\dfrac{-2.0\times10^{-9}}{8.85\times10^{-12}}\right| \)
\( |\Phi_E|=2.26\times10^{2}\,\text{N}\cdot\text{m}^2/\text{C} \)
\( \boxed{|\Phi_E|=226\,\text{N}\cdot\text{m}^2/\text{C}} \)
(b)(i)
For a positive test charge moved slowly from one equipotential to another, the work done by the external force is proportional to the absolute value of the change in electric potential:
\( |W_{\text{ext}}|=q|\Delta V| \)
Since \(D\) and \(E\) are on the same equipotential line,
\( W_{ED}=0 \)
Therefore, the \(W_{ED}\) column should have a \(0\).
The work from \(D\) to \(A\) corresponds to a larger potential difference than the work from \(C\) to \(E\). Since the given \(W_{CE}\) bar is \(4\) units tall, the \(W_{DA}\) bar should be \(6\) units tall.
\( \boxed{W_{ED}=0} \)
\( \boxed{W_{DA}\text{ should be }6\text{ units tall}} \)
(b)(ii)
The electric field component is related to electric potential by
\( E_x=-\dfrac{dV}{dx} \)
An approximate magnitude can be found from nearby equipotential lines:
\( |E_x|\approx \left|\dfrac{\Delta V}{\Delta x}\right| \)
Near Position \(B\), the potential changes from about \(20.0\,\text{V}\) to \(0.0\,\text{V}\) over about \(0.65\,\text{m}\).
\( |E_x|\approx \left|\dfrac{20.0\,\text{V}-0.0\,\text{V}}{0.65\,\text{m}}\right| \)
\( |E_x|\approx 31\,\text{V/m} \)
\( \boxed{|E_x|\approx 31\,\text{V/m}} \)
(c)
Correct direction:
\( \boxed{+y} \)
The electric field is perpendicular to the equipotential line at Position \(D\). A positive test charge experiences an electric force in the same direction as the electric field. Since a positive charge moves from higher electric potential to lower electric potential when released, the force at \(D\) is directed upward, or in the \(+y\)-direction.
(d)(i)
The electric potential due to a small charge element is
\( dV=\dfrac{k\,dq}{r} \)
For the uniformly charged rod,
\( dq=\lambda\,dx \)
A charge element at position \(x\) is a distance
\( r=x_P-x \)
from point \(P\). Therefore,
\( dV=\dfrac{k\lambda\,dx}{x_P-x} \)
The rod extends from \(x=0\) to \(x=4L\), so
\( V_P=\int_0^{4L}\dfrac{k\lambda}{x_P-x}\,dx \)
Factor out the constants:
\( V_P=k\lambda\int_0^{4L}\dfrac{1}{x_P-x}\,dx \)
Since
\( \int \dfrac{1}{x_P-x}\,dx=-\ln|x_P-x| \)
then
\( V_P=k\lambda\left[-\ln|x_P-x|\right]_0^{4L} \)
\( V_P=k\lambda\left[-\ln|x_P-4L|+\ln|x_P|\right] \)
\( V_P=k\lambda\ln\left|\dfrac{x_P}{x_P-4L}\right| \)
\( \boxed{V_P=k\lambda\ln\left|\dfrac{x_P}{x_P-4L}\right|} \)
(d)(ii)
For \(4L<x<12L\), the electric field points in the \(+x\)-direction because the rod is positively charged and the point is to the right of the rod.
The field is strongest just to the right of \(x=4L\), then decreases as \(x\) increases. The graph should therefore be:
• always positive for \(4L<x<12L\)
• very large near \(x=4L\)
• decreasing as \(x\) increases
• concave up, approaching \(0\) as \(x\) gets far from the rod
A consistent expression for the field is
\( E_x=-\dfrac{dV}{dx} \)
or directly from the rod,
\( E_x=k\lambda\left(\dfrac{1}{x-4L}-\dfrac{1}{x}\right) \)
\( E_x=\dfrac{4k\lambda L}{x(x-4L)} \)
This expression is positive and decreases as \(x\) increases, matching the required sketch.