A capacitor with capacitance 5 μ F is charged to 5 μ C. If the plates are pulled apart to reduce the capacitance to 2 1/4 F, how much work is done ?

Question

A capacitor with capacitance 5 μ F is charged to 5 μ C. If the plates are pulled apart to reduce the capacitance to 2 1/4 F, how much work is done ? (A) 3.75 × 10-6 J (B) 2.55 × 10-6 J (C) 2.16 × 10-6 J (D) 6.25 × 10-6 J

Tardigrade Admin · Accepted Answer

Work done = ΔU = Uf - Ui = (q2/2Cr) - (q2/2Ci) = ((5 × 10-6)2/2) ((1/2 × 10-6) - (1/5 × 10-6)) = (15/4) × 10-6 = 3.75 × 10-6J

Q. A capacitor with capacitance $5 μ F$ is charged to $5 μ C$ . If the plates are pulled apart to reduce the capacitance to $2$ $1/4 F$ , how much work is done ?

3.75 × 10 $^{- 6}$ J

2.55 × 10 $^{- 6}$ J

2.16 × 10 $^{- 6}$ J

6.25 × 10 $^{- 6}$ J

Work done = $Δ$ U
$= U_{f} - U_{i}$
$= \frac{q ^{2}}{2 C _{r}} - \frac{q ^{2}}{2 C _{i}}$
$= \frac{( 5 \times 1 0 ^{- 6} ) ^{2}}{2} (\frac{1}{2 \times 1 0 ^{- 6}} - \frac{1}{5 \times 1 0 ^{- 6}})$
$= \frac{15}{4} \times 1 0^{- 6}$
$= 3.75 \times 1 0^{- 6} J$

Q. A capacitor with capacitance 5μF is charged to 5μC. If the plates are pulled apart to reduce the capacitance to 2 1/4F, how much work is done ?

3.75 × 10−6 J

2.55 × 10−6 J

2.16 × 10−6 J

6.25 × 10−6 J