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Question: In the rectangle, shown below, the two corners have charges \(q_{1} = - 5\mu C\) and\(q_{2} = + 2.0\...

In the rectangle, shown below, the two corners have charges q1=−5μCq_{1} = - 5\mu C andq2=+2.0μCq_{2} = + 2.0\mu C. The work done in moving a charge +3.0μC+ 3.0\mu C from BB to AA is

(take1/4πε0=1010N−m2⥂/C21/4\pi\varepsilon_{0} = 10^{10}N - m^{2} ⥂ /C^{2})

A

2.8 J

B

3.5 J

C

4.5 J

D

5.5 J

Answer

2.8 J

Explanation

Solution

Work done W=3×10−6(VA−VB);W = 3 \times 10^{- 6}(V_{A} - V_{B}); where

VA=1010[(−5×10−6)15×10−2+2×10−65×10−2]=115×106voltV_{A} = 10^{10}\left\lbrack \frac{( - 5 \times 10^{- 6})}{15 \times 10^{- 2}} + \frac{2 \times 10^{- 6}}{5 \times 10^{- 2}} \right\rbrack = \frac{1}{15} \times 10^{6}volt

and VB=1010[(2×10−6)15×10−2−5×10−65×10−2]=−1315×106voltV_{B} = 10^{10}\left\lbrack \frac{(2 \times 10^{- 6})}{15 \times 10^{- 2}} - \frac{5 \times 10^{- 6}}{5 \times 10^{- 2}} \right\rbrack = - \frac{13}{15} \times 10^{6}volt

∴ W=3×10−6[115×106−(−1315×106)]W = 3 \times 10^{- 6}\left\lbrack \frac{1}{15} \times 10^{6} - \left( - \frac{13}{15} \times 10^{6} \right) \right\rbrack= 2.8 J