Coulomb's Law — Question 7

Section 21-3 · Physics pages 3424–3425

Question 7 A total charge and force measurement determine an unordered pair of charges Use k = 8.99 × 109 N m2/C2; treat the specified charges as points in vacuum. Two small conducting spheres, far enough apart to use a point approximation, have total charge +10.0 nC. At separation .300 m they repel with force 2.0976667 × 10−6 N (rounded from an exact model value). They are then brought into a symmetric contact arrangement, disconnected, and returned to the same separation. The spheres are identical. A B .300 m Schematic; labels, not drawing lengths, specify the data. (a) Infer the two original charges, retaining the two possible assignments to labeled spheres. (b) Find their final charges and the ratio of final to initial force. Explain why sharing increases the force for these same-sign charges.
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Question 7 - Solution A total charge and force measurement determine an unordered pair of charges A B .300 m (a) Method and calculation qA + qB = 10 nC, qAqB = Fr2/k ≈21 nC2. z2 −10z + 21 = 0 ⇒z = 3 or 7, {qA, qB} = {3, 7} nC. (b) Interpretation and further calculation qA′ = qB′ = 5 nC, F′/F = 25/21 ≈1.19048. qAqB = (qA + qB)2 −(qA −qB)2 4 ≤(qA + qB)2 4 . Equal sharing maximizes the product at fixed sum; the initial measurements do not identify which sphere originally carried 3 nC. Physical check The rounded measured force supports approximately 3 and 7 nC; it does not justify more digits than the input precision.

Question 7

A total charge and force measurement determine an unordered pair of charges

Use k=8.99×109Nm2/C2k=8.99\times10^9\,\mathrm{N\,m^2/C^2}; treat the specified charges as points in vacuum. Two small conducting spheres, far enough apart to use a point approximation, have total charge +10.0+10.0 nC. At separation .300 m they repel with force 2.0976667×10−62.0976667\times10^{-6} N (rounded from an exact model value). They are then brought into a symmetric contact arrangement, disconnected, and returned to the same separation. The spheres are identical.

Diagram shown in the original SVG worksheet.

  1. Infer the two original charges, retaining the two possible assignments to labeled spheres.

  2. Find their final charges and the ratio of final to initial force. Explain why sharing increases the force for these same-sign charges.

Question 7 - Solution

A total charge and force measurement determine an unordered pair of charges

Diagram shown in the original SVG worksheet.

(a) Method and calculation

qA+qB=10nC,qAqB=Fr2/k≈21nC2.q_A+q_B=10\,\mathrm{nC},\quad q_Aq_B=Fr^2/k\approx21\,\mathrm{nC^2}.z2−10z+21=0⇒z=3 or 7,{qA,qB}={3,7}nC.z^2-10z+21=0\Rightarrow z=3\text{ or }7,\quad\{q_A,q_B\}=\{3,7\}\,\mathrm{nC}.

(b) Interpretation and further calculation

qA′=qB′=5nC,F′/F=25/21≈1.19048.q_A\prime=q_B\prime=5\,\mathrm{nC},\quad F\prime/F=25/21\approx1.19048.qAqB=(qA+qB)2−(qA−qB)24≤(qA+qB)24.q_Aq_B=\frac{(q_A+q_B)^2-(q_A-q_B)^2}{4}\le\frac{(q_A+q_B)^2}{4}.Equal sharing maximizes the product at fixed sum; the initial measurements do not identify which sphere originally carried 3 nC.

Physical check

The rounded measured force supports approximately 3 and 7 nC; it does not justify more digits than the input precision.