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Capacitor question

2022 · 25 Jul · Shift 1 · Q50
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  5. /2022 · 25 Jul · Shift 1 · Q50

Capacitor question

2022 · 25 Jul · Shift 1 · Q50

JEE MainPhysicsCapacitorMCQ+4 / −1
A condenser of 2 μF2 \,\mu \mathrm{F}2μF capacitance is charged steadily from 0 to 5 C5 \,\mathrm{C}5C. Which of the following graph represents correctly the variation of potential difference (V)(\mathrm{V})(V) across it's plates with respect to the charge (Q)(Q)(Q) on the condenser?
  1. A
    JEE Main 2022 (Online) 25th July Morning Shift Physics - Capacitor Question 60 English Option 1
  2. B
    JEE Main 2022 (Online) 25th July Morning Shift Physics - Capacitor Question 60 English Option 2
  3. C
    JEE Main 2022 (Online) 25th July Morning Shift Physics - Capacitor Question 60 English Option 3
  4. D
    JEE Main 2022 (Online) 25th July Morning Shift Physics - Capacitor Question 60 English Option 4
View written solutionFree

Correct answer: A

  1. For a capacitor, the relation between charge and potential difference is

Q=CVQ = CVQ=CV

So,

V=QCV = \frac{Q}{C}V=CQ​

  1. Here, the capacitance is constant:

C=2 μF=2×10−6 FC = 2\,\mu\text{F} = 2 \times 10^{-6}\,\text{F}C=2μF=2×10−6F

Hence,

V=Q2×10−6V = \frac{Q}{2\times 10^{-6}}V=2×10−6Q​

This shows that VVV is directly proportional to QQQ.

  1. Therefore, the graph of VVV versus QQQ must be:
  • a straight line
  • passing through the origin
  • with positive slope
  1. The slope of the graph is

VQ=1C=12×10−6=5×105 V/C\frac{V}{Q} = \frac{1}{C} = \frac{1}{2\times 10^{-6}} = 5\times 10^5\,\text{V/C}QV​=C1​=2×10−61​=5×105V/C

So as charge increases linearly, potential difference also increases linearly.

  1. Therefore, the correct graph is the one showing a straight line through the origin.

Thus, the correct option is A.

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