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Fundamental and Derived Units

Charges 2μ ...

Question

Charges $2 μ C$ and $1 μ C$ are placed at corners $A$ and $B$ of square of side $5 c m$ as shown in figure. The amount of work to be done against the electric field in moving a charge of $1.0 \times 10^{- 6} C$ from $C$ to $D$ is :

0.045 J

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0.053 J

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0.017 J

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0.03 J

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Solution

The correct option is B $0.053 J$

Work to be done against the electric field in electric field in moving a charge $= 1.0 \times 10^{- 6} C$
$F = \frac{1}{4 π ϵ_{0}} \times \frac{q_{1} q_{2}}{r^{2}}$
$= 9 \times 10^{9} \times \frac{2 \times 10^{- 7} \times 3 \times 10^{- 7}}{{(5 \times 10^{- 2})}^{2}}$
$= 2.16 \times 10^{- 19} N$
$W = J F$
or, $J = \frac{W}{F}$
$E = \frac{F}{q} = \frac{2.16 \times 10^{- 19}}{1.0 \times 10^{- 6}} = 0.053 J$

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Charges 2μC and 1μC are placed at corners A and B of square of side 5cm as shown in figure. The amount of work to be done against the electric field in moving a charge of 1.0×10−6C from C to D is :

The correct option is B 0.053JWork to be done against the electric field in electric field in moving a charge =1.0×10−6CF=14πϵ0×q1q2r2 =9×109×2×10−7×3×10−7(5×10−2)2 =2.16×10−19NW=JFor, J=WFE=Fq=2.16×10−191.0×10−6=0.053J

Charges $2 μ C$ and $1 μ C$ are placed at corners $A$ and $B$ of square of side $5 c m$ as shown in figure. The amount of work to be done against the electric field in moving a charge of $1.0 \times 10^{- 6} C$ from $C$ to $D$ is :