Question

In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q₂ as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)

Solution

From the given circuit,
Q_c= Q₁+Q₂
or
C (E-V) = 1 x V + 2 XV
Or V (C+3) = CE
$straight V space equals space fraction numerator CE over denominator 3 plus straight C end fraction straight Q subscript 2 space equals space straight C subscript 2 space left parenthesis straight V right parenthesis space equals space fraction numerator 2 CE over denominator 3 plus straight C end fraction equals space fraction numerator 2 straight E over denominator 1 plus 3 divided by straight C end fraction$
As, C₁ varied from 1μF to 3μF, charge increases with decreasing slope.

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Electrostatic Potential And Capacitance

In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q₂ as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)

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Electrostatic Potential And Capacitance

In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q2 as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)

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In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q₂ as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)