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ExamsJEE AdvancedPhysics

A parallel-plate capacitor of capacitance C is connected to a battery and charged to a potential difference V. With the battery still connected, the separation between the plates is slowly reduced to half its original value. What is the work done by the external agent during this process?

  1. 1/2 * C * V²
  2. C * V²
  3. -1/2 * C * V²
  4. -C * V²

Correct answer: -1/2 * C * V²

Solution

With V fixed, new capacitance is 2C. Energy stored increases from (1/2)CV² to CV². Battery does work CV² (supplies extra charge CV at potential V). By energy conservation, W_external = delta_U - W_battery = (1/2)CV² - CV² = -1/2*CV².

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