A) decreases
B) does not change
C) becomes zero
D) increases
Correct Answer: D
Solution :
| [d] Key Idea: Charge remains constant after charging. |
| If the battery is removed after charging then the charge stored in the capacitor remains constant. |
| \[q=\]constant |
| Change in capacitance |
| \[C'=\frac{{{\varepsilon }_{0}}\,A}{d'}\] |
| As \[d'>d\] |
| hence, \[C'<C\] |
| Hence, potential difference between the plates |
| or \[V'\,\,\propto \,\,\frac{1}{C'}\] |
| As capacitance decreases, so potential difference increases. |
| Note: If the battery remains connected, the charge stores increases. Also the potential difference V becomes constant. |
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