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CBSE-XI-Physics

31: Capacitors

with Solutions - page 3
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  • #5
    A parallel-plate capacitor has plate area 25⋅0 cm2 and a separation of 2⋅00 mm between the plates. The capacitor is connected to a battery of 12⋅0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate.
    Ans : Given:
    Area of the plate, A = 25 cm2 = 25 × 10`` -``4 m2
    Separation between the plates, d = 2 mm = 2 × 10`` -``3 m
    Potential difference between the plates, V = 12 V
    The capacitance of the given capacitor is given by
    `` C=\frac{{\in }_{0}A}{d}``
    `` =\frac{(8.85\times {10}^{-12})\times (25\times {10}^{-4})}{(2\times {10}^{-3})}``
    `` =11.06\times {10}^{-12}\,\mathrm{\,F\,}`` (a) Charge on the capacitor is given by
    Q = CV
    `` =11.06\times {10}^{-12}\times 12``
    `` =1.33\times {10}^{-10}\,\mathrm{\,C\,}`` (b) When the separation between the plates is decreased to 1 mm, the capacitance C' can be calculated as:
    `` C\text{'}=\frac{{\in }_{0}\,\mathrm{\,A\,}}{d}``
    `` =\frac{(8.85\times {10}^{-12})\times (25\times {10}^{-4})}{1\times {10}^{-3}}``
    `` =22.12\times {10}^{-12}\,\mathrm{\,F\,}``
    `` ``
    Charge on the capacitor is given by
    Q' = C'V
    `` =22.12\times {10}^{-12}\times 12``
    `` =2.65\times {10}^{-10}\,\mathrm{\,C\,}``
    `` \,\mathrm{\,Extra\,}\,\mathrm{\,charge\,}``
    `` =\left(2.65\times {10}^{-10}-1.32\times {10}^{-10}\right)\,\mathrm{\,C\,}``
    `` =1.33\times {10}^{-10}\,\mathrm{\,C\,}``
    Page No 165: (a) Charge on the capacitor is given by
    Q = CV
    `` =11.06\times {10}^{-12}\times 12``
    `` =1.33\times {10}^{-10}\,\mathrm{\,C\,}`` (b) When the separation between the plates is decreased to 1 mm, the capacitance C' can be calculated as:
    `` C\text{'}=\frac{{\in }_{0}\,\mathrm{\,A\,}}{d}``
    `` =\frac{(8.85\times {10}^{-12})\times (25\times {10}^{-4})}{1\times {10}^{-3}}``
    `` =22.12\times {10}^{-12}\,\mathrm{\,F\,}``
    `` ``
    Charge on the capacitor is given by
    Q' = C'V
    `` =22.12\times {10}^{-12}\times 12``
    `` =2.65\times {10}^{-10}\,\mathrm{\,C\,}``
    `` \,\mathrm{\,Extra\,}\,\mathrm{\,charge\,}``
    `` =\left(2.65\times {10}^{-10}-1.32\times {10}^{-10}\right)\,\mathrm{\,C\,}``
    `` =1.33\times {10}^{-10}\,\mathrm{\,C\,}``
    Page No 165:
  • #5-a
    Find the charge on the capacitor.
    Ans : Charge on the capacitor is given by
    Q = CV
    `` =11.06\times {10}^{-12}\times 12``
    `` =1.33\times {10}^{-10}\,\mathrm{\,C\,}``
  • #5-b
    The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate.
    Ans : When the separation between the plates is decreased to 1 mm, the capacitance C' can be calculated as:
    `` C\text{'}=\frac{{\in }_{0}\,\mathrm{\,A\,}}{d}``
    `` =\frac{(8.85\times {10}^{-12})\times (25\times {10}^{-4})}{1\times {10}^{-3}}``
    `` =22.12\times {10}^{-12}\,\mathrm{\,F\,}``
    `` ``
    Charge on the capacitor is given by
    Q' = C'V
    `` =22.12\times {10}^{-12}\times 12``
    `` =2.65\times {10}^{-10}\,\mathrm{\,C\,}``
    `` \,\mathrm{\,Extra\,}\,\mathrm{\,charge\,}``
    `` =\left(2.65\times {10}^{-10}-1.32\times {10}^{-10}\right)\,\mathrm{\,C\,}``
    `` =1.33\times {10}^{-10}\,\mathrm{\,C\,}``
    Page No 165: