# Answer in Electric Circuits

October 16th, 2022

**Explanations & Calculations**

- A Capacitor, when connected to a voltage source, stores electrical charges on both its conducting plates until the potential difference between those plates get equal to that of the external source.
- For a given capacitance & a voltage source, it stores a fixed amount of charges according to the equation “small Q=CV”
- Once the voltage source is removed, the capacitor retains those stored charges on it & the potential difference between the plates still remains in the previous value.
- According to the definition of the capacitance—how many charges stores under a unit voltage—what happens after capacitance decreased is that those stored charges equivalents to some higher potential difference or that much of charges could be stored in that capacitor under comparatively higher potential difference.
- Energy stored in a capacitor is given by “small W=largefrac{1}{2}frac{Q^2}{C}”. According to that energy stored in it increases as the capacitance is decreased.
- The assumption made here is “no charge loss”.

1)

- Charges stored in the capacitor in the first arrangement,

“qquadqquadnbegin{aligned}nsmall Q_i&= C_iV_i\n&= small (5times10^{-10}F)(200V)\n&=small 1times10^{-7},Cnend{aligned}”

2)

- After the capacitance decreased,new potential difference would be

“qquadqquadnbegin{aligned}nsmall Q_i&= small C_{new}V_{new}\nsmall 1times10^{-7}&= small (1times10^{-10}F).V_{new}\nsmall V_{new}&= small 10^3\nsmall V_{new}&=small bold{1000V} nend{aligned}”

3)

- Work needs to be done to change the capacitance is

“qquadqquadnbegin{aligned}nsmall Delta W&= small W_{new}-W_i\n&= small frac{1}{2}frac{Q^2}{C_{new}}-frac{1}{2}frac{Q^2}{C_i}\n&= small frac{Q^2}{2}bigg[frac{1}{C_{new}}-frac{1}{C_i}bigg] \n&= small frac{(1times10^{-7}C)^2}{2}bigg[frac{1}{1times10^{-10}F}-frac{1}{5times10^{-10}F}bigg]\n&=smallbold{4times10^{-5}J}nend{aligned}”

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