IB Physics HL Theme D: Fields questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Multiple choiceMedium1 mark

A satellite is at a distance r=2.0×107 mr=2.0\times10^7\text{ m} from the centre of a planet. The graph shows the planet’s gravitational field strength gg at selected distances rr What is the escape speed from this position?

0123405101520r/(107 m)g/(N kg-1)
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Answer: B. At r=2.0×107 mr=2.0\times10^7\text{ m} the graph gives g=5.0 N kg−1g=5.0\text{ N kg}^{-1} Since g=GM/r2g=GM/r^2 vesc=2GM/r=2gr=2(5.0)(2.0×107)=1.4×104 m s−1v_{\rm esc}=\sqrt{2GM/r}=\sqrt{2gr}=\sqrt{2(5.0)(2.0\times10^7)}=1.4\times10^4\text{ m s}^{-1}

A: uses the circular orbital speed gr\sqrt{gr} omitting the factor 2 under the square root.
C: assumes escape speed is twice the circular orbital speed, rather than 2\sqrt{2} times it.
D: reads g=20 N kg−1g=20\text{ N kg}^{-1} from the graph at r=1.0×107 mr=1.0\times10^7\text{ m} not the value at the satellite’s radius.

Remember: escape speed is 2\sqrt{2} times circular orbital speed at the same radius.

Question 2

Multiple choiceMedium1 mark

Two fixed point charges, +Q+Q and −Q-Q are separated by 2a2a where Q>0Q>0 Point PP is midway between them. Let kk be Coulomb’s constant. Which row gives the electric potential at PP and the location chosen as the zero of electric potential?

Electric potential at PPZero-potential reference
A.00infinity
B.2kQa\dfrac{2kQ}{a}infinity
C.00PP
D.kQa\dfrac{kQ}{a}infinity
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Answer: A. For a point charge, V=kq/rV=kq/r when the potential is zero at infinity. At PP the contributions are +kQ/a+kQ/a and −kQ/a-kQ/a so their scalar sum is V(P)=0V(P)=0 The reference location for zero potential is infinity.

B: adds the magnitudes of the two contributions, ignoring the negative charge’s sign.
C: confuses a point where the net potential is zero with the location chosen as the zero reference.
D: includes the contribution from +Q+Q but omits the equal contribution from −Q-Q

Remember: electric potentials add as signed scalars.

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