Q166 · Practice questionComplex unfamiliar5 marks
QUESTION 166 (5 marks)
On the -axis, a fixed charge is at and a fixed charge is at . Consider finite points on this axis, excluding the charge positions.
a)[2 marks]
Explain why the net electric field cannot be zero between the charges or to the right of both charges.
b)[3 marks]
Determine the coordinate of the finite field-cancellation point.
WORKED SOLUTION
5 marksPractice marking scheme
ANSWER
(a) Between them both fields point right. To their right the larger, nearer negative charge gives the stronger field. (b) .
Worked solution
(a) Between the charges the field points away from the positive charge and toward the negative charge, both rightward. To the right of both, the negative charge has greater magnitude and is also closer, so its leftward field cannot be balanced by the weaker rightward field.
(b) Let to the left of both charges. Set . Taking positive roots gives , so , and .
Equivalent physically justified methods and consistent equivalent units accepted.
Displayed decimals are model answers; accept appropriate significant figures and consistent rounding from stated constants.
Part a: Identify reinforcing fields between the charges.
Part a: Explain the dominance of the negative charge to the right.
Part b: Equate the magnitudes of the opposing fields to the left.
Part b: Solve using the unequal distances to the charges.
Part b: State the signed coordinate.
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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