184_notes:examples:week8_cap_series

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 184_notes:examples:week8_cap_series [2021/06/28 20:03]schram45 184_notes:examples:week8_cap_series [2021/06/28 20:08] (current)schram45 [Solution] 2021/06/28 20:08 schram45 [Solution] 2021/06/28 20:03 schram45 2021/06/28 20:01 schram45 2021/06/21 14:22 schram45 2018/06/26 10:39 curdemma [Solution] 2018/06/26 10:39 curdemma 2017/10/16 19:50 dmcpadden 2017/10/11 09:53 tallpaul [Example: Resistors in Series] 2017/10/11 09:42 tallpaul 2017/10/11 08:56 tallpaul [Example: Resistors in Series] 2017/10/11 08:20 tallpaul 2017/10/10 09:44 tallpaul created 2021/06/28 20:08 schram45 [Solution] 2021/06/28 20:03 schram45 2021/06/28 20:01 schram45 2021/06/21 14:22 schram45 2018/06/26 10:39 curdemma [Solution] 2018/06/26 10:39 curdemma 2017/10/16 19:50 dmcpadden 2017/10/11 09:53 tallpaul [Example: Resistors in Series] 2017/10/11 09:42 tallpaul 2017/10/11 08:56 tallpaul [Example: Resistors in Series] 2017/10/11 08:20 tallpaul 2017/10/10 09:44 tallpaul created Line 47: Line 47: **This is the charge on __both__ capacitors** since the capacitors are in series. So even if we insert a dielectric in only one of the capacitors, the charge on both will increase. **This is the charge on __both__ capacitors** since the capacitors are in series. So even if we insert a dielectric in only one of the capacitors, the charge on both will increase. + + In order for us to evaluate this solution and make sure it makes sense, we must understand what the dielectric is doing in the capacitor. The dielectric is an insulator which means its electrons are tied to the nuclei. When a dielectric is put between two charged plates it will polarize. This will create an electric field in the dielectric that opposes the electric field of the plates. This will allow more charges to be added to the plates before the capacitor will have an electric field that fully opposes that of the power source. This solution lines up with that perfectly.
• 184_notes/examples/week8_cap_series.txt