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When 625×10^16 electrons are removed from a spherical conductor, the potential at the surface of the conductor is found to be 1 V . The capacitance of the conductor is (A) 1F. (B) 0.1F. (C) 1mF. (D) 1×10^(-6) F?
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When 625×10^16 electrons are removed from a spherical conductor, the p...
Given Data:
- Number of electrons removed, n = 625×10^16
- Potential at the surface, V = 1 V

Capacitance Calculation:
- The charge on the conductor, Q = n×e (where e is the charge of an electron)
- From the formula for capacitance, C = Q/V, we can find the capacitance of the conductor.

Calculation:
- Charge of 1 electron, e = 1.6×10^(-19) C
- Total charge, Q = n×e = 625×10^16 × 1.6×10^(-19) = 1×10^(-2) C
- Capacitance, C = Q/V = 1×10^(-2) / 1 = 1×10^(-2) F

Conclusion:
- The capacitance of the conductor is 1×10^(-2) F, which is equivalent to 10mF.
- Therefore, the correct option is (C) 1mF.
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When 625×10^16 electrons are removed from a spherical conductor, the potential at the surface of the conductor is found to be 1 V . The capacitance of the conductor is (A) 1F. (B) 0.1F. (C) 1mF. (D) 1×10^(-6) F?
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When 625×10^16 electrons are removed from a spherical conductor, the potential at the surface of the conductor is found to be 1 V . The capacitance of the conductor is (A) 1F. (B) 0.1F. (C) 1mF. (D) 1×10^(-6) F? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about When 625×10^16 electrons are removed from a spherical conductor, the potential at the surface of the conductor is found to be 1 V . The capacitance of the conductor is (A) 1F. (B) 0.1F. (C) 1mF. (D) 1×10^(-6) F? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 625×10^16 electrons are removed from a spherical conductor, the potential at the surface of the conductor is found to be 1 V . The capacitance of the conductor is (A) 1F. (B) 0.1F. (C) 1mF. (D) 1×10^(-6) F?.
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