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Question: Two point charge 100 *μC* and 5 *μC* are placed at point *A* and *B* respectively with *AB* = 40 *cm...

Two point charge 100 μC and 5 μC are placed at point A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 μC from B to C, where BC = 30 cm, angle ABC=π2and 14πε0=9×109Nm2⥂/⥂C2ABC = \frac{\pi}{2}\text{and }\frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}Nm^{2} ⥂ / ⥂ C^{2}

A

9 J

B

8120J\frac{81}{20}J

C

925J\frac{9}{25}J

D

−94J- \frac{9}{4}J

Answer

−94J- \frac{9}{4}J

Explanation

Solution

Potential at B due to +100 μC charge is

VB=9×109×100×10−640×10−2=94×106voltV_{B} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{40 \times 10^{- 2}} = \frac{9}{4} \times 10^{6}voltPotential at C due

to +100 μC charge is

VC=9×109×100×10−650×10−2=95×106voltV_{C} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{50 \times 10^{- 2}} = \frac{9}{5} \times 10^{6}voltHence work done in moving charge +5μC from B to C

W=5×10−6(VC−VB)W = 5 \times 10^{- 6}(V_{C} - V_{B})

W=5×10−6(95×106−94×10+6)=−94JW = 5 \times 10^{- 6}\left( \frac{9}{5} \times 10^{6} - \frac{9}{4} \times 10^{+ 6} \right) = - \frac{9}{4}J

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