SATHEE: Electrostatics Question 377

Electrostatics Question 377

Question: Two point charges $100 μ C$ and $5 μ C$ are placed at points $A$ and $B$ respectively with $A B = 40 c m$ . The work done by external force in displacing the charge $5 μ C$ from $B$ to $C$ , where $B C = 30 c m$ , angle $A B C = \frac{π}{2}$ and $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} / C^{2}$ [MP PMT 1997]

Options:

A) $9 J$

B) $\frac{81}{20} J$

C) $\frac{9}{25} J$

D) $- \frac{9}{4} J$

Show Answer

Answer:

Correct Answer: D

Solution:

Work done in displacing charge of 5 m C from B to C is $W = 5 \times 10^{- 6} (V_{C} - V_{B})$

where $V_{B} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0.4} = \frac{9}{4} \times 10^{6} V$

and $V_{C} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0.5} = \frac{9}{5} \times 10^{6} V$

So $W = 5 \times 10^{- 6} \times (\frac{9}{5} \times 10^{6} - \frac{9}{4} \times 10^{6}) = - \frac{9}{4} J$

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Electrostatics Question 377

Question: Two point charges 100μC and 5μC are placed at points A and B respectively with AB=40cm . The work done by external force in displacing the charge 5μC from B to C , where BC=30cm , angle ABC=π2 and 14πε0=9×109Nm2/C2 [MP PMT 1997]

Options:

Answer:

Solution:

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