JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 10)
A condenser of $$2 \,\mu \mathrm{F}$$ capacitance is charged steadily from 0 to $$5 \,\mathrm{C}$$. Which of the following graph represents correctly the variation of potential difference $$(\mathrm{V})$$ across it's plates with respect to the charge $$(Q)$$ on the condenser?
_25th_July_Morning_Shift_en_10_1.png)
_25th_July_Morning_Shift_en_10_2.png)
_25th_July_Morning_Shift_en_10_3.png)
_25th_July_Morning_Shift_en_10_4.png)
Explanation
Q = CV
As capacitance is constant so, Q $$\propto$$ V
So graph between V and Q will be a straight line.
Initially voltage, $${V_i} = {{{Q_i}} \over C} = {0 \over {2 \times {{10}^{ - 6}}}} = 0 V$$
and $${V_f} = {{{Q_f}} \over C} = {5 \over {2 \times {{10}^{ - 6}}}} = 2.5 \times {10^6}\,V$$
So correct graph will be A.
Comments (0)
