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Electrostatics question

2022 · 26 Jul · Shift 2 · Q63
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  5. /2022 · 26 Jul · Shift 2 · Q63

Electrostatics question

2022 · 26 Jul · Shift 2 · Q63

JEE MainPhysicsElectrostaticsNumerical+4 / −1
Three point charges of magnitude 5μC,0.16μC5 \mu \mathrm{C}, 0.16 \mu \mathrm{C}5μC,0.16μC and 0.3μC0.3 \mu \mathrm{C}0.3μC are located at the vertices A,B,CA, B, CA,B,C of a right angled triangle whose sides are AB=3 cm,BC=32 cmA B=3 \mathrm{~cm}, B C=3 \sqrt{2} \mathrm{~cm}AB=3 cm,BC=32​ cm and CA=3 cmC A=3 \mathrm{~cm}CA=3 cm and point AAA is the right angle corner. Charge at point A\mathrm{A}A experiences ‾\underline{\hspace{2cm}}​N\mathrm{N}N of electrostatic force due to the other two charges.
Numerical answer
View written solutionFree

Correct answer: 17

  1. Understand the geometry

Given triangle ABCABCABC is right-angled at AAA with AB=3 cm,AC=3 cm,BC=32 cmAB=3\text{ cm},\quad AC=3\text{ cm},\quad BC=3\sqrt{2}\text{ cm}AB=3 cm,AC=3 cm,BC=32​ cm So AB⊥ACAB \perp ACAB⊥AC.

Thus, the charge at AAA experiences:

  • force due to charge at BBB along ABABAB
  • force due to charge at CCC along ACACAC

These two forces are perpendicular.

  1. Assign charges

Magnitude of charges at the vertices are: qA=5 μC,qB=0.16 μC,qC=0.3 μCq_A=5\,\mu C,\quad q_B=0.16\,\mu C,\quad q_C=0.3\,\mu CqA​=5μC,qB​=0.16μC,qC​=0.3μC

We need the net electrostatic force on charge at AAA due to charges at BBB and CCC.

Convert to SI units: qA=5×10−6 Cq_A=5\times 10^{-6}\,CqA​=5×10−6C qB=0.16×10−6 Cq_B=0.16\times 10^{-6}\,CqB​=0.16×10−6C qC=0.3×10−6 Cq_C=0.3\times 10^{-6}\,CqC​=0.3×10−6C AB=AC=3 cm=0.03 mAB=AC=3\text{ cm}=0.03\text{ m}AB=AC=3 cm=0.03 m

Use Coulomb's law: F=k∣q1q2∣r2,k=9×109 N m2/C2F = k\frac{|q_1q_2|}{r^2}, \qquad k=9\times 10^9\,\text{N m}^2/\text{C}^2F=kr2∣q1​q2​∣​,k=9×109N m2/C2

  1. Force on charge at AAA due to charge at BBB

FAB=kqAqBAB2F_{AB}=k\frac{q_Aq_B}{AB^2}FAB​=kAB2qA​qB​​

Substitute: FAB=9×109⋅(5×10−6)(0.16×10−6)(0.03)2F_{AB}=9\times 10^9\cdot \frac{(5\times 10^{-6})(0.16\times 10^{-6})}{(0.03)^2}FAB​=9×109⋅(0.03)2(5×10−6)(0.16×10−6)​

First compute numerator: (5×10−6)(0.16×10−6)=0.8×10−12(5\times 10^{-6})(0.16\times 10^{-6})=0.8\times 10^{-12}(5×10−6)(0.16×10−6)=0.8×10−12

So, FAB=9×109⋅0.8×10−129×10−4F_{AB}=9\times 10^9\cdot \frac{0.8\times 10^{-12}}{9\times 10^{-4}}FAB​=9×109⋅9×10−40.8×10−12​

=7.2×10−39×10−4=8 N=\frac{7.2\times 10^{-3}}{9\times 10^{-4}}=8\,\text{N}=9×10−47.2×10−3​=8N

So, FAB=8 NF_{AB}=8\,\text{N}FAB​=8N

  1. Force on charge at AAA due to charge at CCC

FAC=kqAqCAC2F_{AC}=k\frac{q_Aq_C}{AC^2}FAC​=kAC2qA​qC​​

Substitute: FAC=9×109⋅(5×10−6)(0.3×10−6)(0.03)2F_{AC}=9\times 10^9\cdot \frac{(5\times 10^{-6})(0.3\times 10^{-6})}{(0.03)^2}FAC​=9×109⋅(0.03)2(5×10−6)(0.3×10−6)​

First compute numerator: (5×10−6)(0.3×10−6)=1.5×10−12(5\times 10^{-6})(0.3\times 10^{-6})=1.5\times 10^{-12}(5×10−6)(0.3×10−6)=1.5×10−12

Thus, FAC=9×109⋅1.5×10−129×10−4F_{AC}=9\times 10^9\cdot \frac{1.5\times 10^{-12}}{9\times 10^{-4}}FAC​=9×109⋅9×10−41.5×10−12​

=13.5×10−39×10−4=15 N=\frac{13.5\times 10^{-3}}{9\times 10^{-4}}=15\,\text{N}=9×10−413.5×10−3​=15N

So, FAC=15 NF_{AC}=15\,\text{N}FAC​=15N

  1. Resultant force at AAA

Since AB⊥ACAB \perp ACAB⊥AC, the forces are perpendicular. Hence net force is F=FAB2+FAC2F=\sqrt{F_{AB}^2+F_{AC}^2}F=FAB2​+FAC2​​

F=82+152F=\sqrt{8^2+15^2}F=82+152​ F=64+225F=\sqrt{64+225}F=64+225​ F=289=17 NF=\sqrt{289}=17\,\text{N}F=289​=17N

  1. Final answer

The charge at point AAA experiences 17 N\boxed{17\,\text{N}}17N​ of electrostatic force.

This matches the stored correct answer.

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