SATHEE: Electrostatics Question 427

Electrostatics Question 427

Question: In the rectangle, shown below, the two corners have charges $q_{1} = - 5 μ C$ and $q_{2} = + 2.0 μ C$ . The work done in moving a charge $+ 3.0 μ C$ from $B$ to $A$ is (take $1 / 4 π ε_{0} = 10^{10} N - m^{2} / C^{2}$ ) [AMU 2001]

Options:

A) 2.8 J

B) 3.5 J

C) 4.5 J

D) 5.5 J

Show Answer

Answer:

Correct Answer: A

Solution:

Work done $W = 3 \times 10^{- 6} (V_{A} - V_{B});$

where $V_{A} = 10^{10} [\frac{(- 5 \times 10^{- 6})}{15 \times 10^{- 2}} + \frac{2 \times 10^{- 6}}{5 \times 10^{- 2}}] = \frac{1}{15} \times 10^{6} v o l t$

and $V_{B} = 10^{10} [\frac{(2 \times 10^{- 6})}{15 \times 10^{- 2}} - \frac{5 \times 10^{- 6}}{5 \times 10^{- 2}}] = - \frac{13}{15} \times 10^{6} v o l t$

$W = 3 \times 10^{- 6} [\frac{1}{15} \times 10^{6} - (- \frac{13}{15} \times 10^{6})]$ = 2.8 J

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Electrostatics Question 427

Question: In the rectangle, shown below, the two corners have charges q1=−5μC and q2=+2.0μC . The work done in moving a charge +3.0μC from B to A is (take 1/4πε0=1010N-m2/C2 ) [AMU 2001]

Options:

Answer:

Solution:

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Question: In the rectangle, shown below, the two corners have charges $q_{1} = - 5 μ C$ and $q_{2} = + 2.0 μ C$ . The work done in moving a charge $+ 3.0 μ C$ from $B$ to $A$ is (take $1 / 4 π ε_{0} = 10^{10} N - m^{2} / C^{2}$ ) [AMU 2001]