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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance4 Marks
Question
A capacitor is a device to store energy. The process of charging up a capacitor involves the transferring of electric charges from its one place to another. This work done in charging the capacitor is stored as its electrical potential energy.

If q is the charge and V is the potential difference across a capacitor at any instant during its charging, then small work done in storing an additional small charge dq against the repulsion of charge q already stored on it is $\text{dW}=\text{V.dq}=(\frac{\text{q}}{\text{C}})\text{dq}.$

  1. A system of 2 capacitors of capacitance $2\mu\text{F}$ and $4\mu\text{F}$ is connected in series across a potential difference of 6 V. The energy stored in the system is:
  1. $3\mu\text{J}$
  2. $24\mu\text{J}$
  3. $30\mu\text{J}$
  4. $108\mu\text{J}$
  1. A capacitor of capacitance of $10\mu\text{F}$ is charged to 10V. The energy stored in it is:
  1. $100\mu\text{J}$
  2. $500\mu\text{J}$
  3. $1000\mu\text{J}$
  4. $1\mu\text{J}$
  1. A parallel plate air capacitor has capacity C farad, potential V volt and energy E joule. When the gap between the plates is completely filled with dielectric:
  1. Both V and E increase.
  2. Both V and E decrease.
  3. V decreases, E increases.
  4. V increases, E decreases.
  1. A capacitor with capacitance $5\mu\text{F}$ s charged to $5\mu\text{C}.$ If the plates are pulled apart to reduce the capacitance to $2\mu\text{F},$ how much work is done?
  1. 6.25 × 10-6J
  2. 3.75 × 10-6J
  3. 2.16 × 10-6J
  4. 2.55 × 10-6J
  1. A metallic sphere ofradius 18cm has been given a charge of 5 × 10-6C. The energy of the charged conductor is:
  1. 0.2J
  2. 0.6J
  3. 1.2J
  4. 2.4J
✓

Answer

  1. (b) $24\mu\text{J}$

Explanation:

As, $\text{C}_1=2\mu\text{F},\text{C}_2=4\mu\text{F}$

ln series combination, the equivalent capacitance will be, 

$\text{C}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\Big(\frac{2\times4}{2+4}\Big)\mu\text{F}$

$=\frac{4}{3}\mu\text{F}$

Potential difference applied, V = 6V

Energy stored in the system, $\text{U}=\frac{1}{2}\text{CV}^2$

$=\frac{1}{2}\times\frac{4}{3}\times10^{-6}\times(6)^2$

$\text{J}=24\mu\text{J}.$

  1. (b) $500\mu\text{J}$

​​​​​​​​​​​​​​Explanation:

The energy stored in a capacitor is

$\text{U}=\frac{1}{2}\text{CV}^2$

$=\frac{1}{2}\times(10\times10^{-6})(10)^2=500\mu\text{J}.$

  1. (b) Both V and E decrease.

​​​​​​​​​​​​​​​​​​​​​Explanation:

When the gap between the plates is completely filled with dielectric of dielectric constant K, then potential is:

$\text{V}=\frac{\text{Qd}}{\text{A}\in_0\text{K}}$ (i)

and electric field is

$\text{E}=\frac{\text{Q}}{\text{A}\in_0\text{K}}$ (ii)

From equations (i) and (ii), both electric field and potential decrease.

  1. (b) 3.75 × 10-6J

​​​​​​​​​​​​​​​​​​​​​​​​​​​​Explanation:

Work done $=\text{U}_\text{f}-\text{U}_\text{i}=\frac{1}{2}\frac{\text{q}^2}{\text{C}_\text{f}}-\frac{1}{2}\frac{\text{q}^2}{\text{C}_\text{i}}$

$=\frac{\text{q}^2}{2}\Big[\frac{1}{\text{C}_\text{f}}-\frac{1}{\text{C}_\text{i}}\Big]$

$=\frac{(5\times10^{-6})^2}{2}\Big[\frac{1}{2\times10^{-6}}-\frac{1}{5\times10^{-6}}\Big]$

= 3.75 × 10-6J

  1. (b) 0.6J

​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Explanation:

Here r = 18cm = 18 × 10-2m, q = 5 × 10-6C

As, $\text{C}=4\pi\in_0\text{r}=\frac{18\times10^{-2}}{9\times10^9}=2\times10^{-11}\text{F}$

Energy of charged conductor is

$\text{U}=\frac{\text{q}^2}{\text{2C}}=\frac{(5\times10^{-6})^2\text{C}}{2\times2\times10^{-11}\text{F}}=0.625\text{J}.$

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