Electrostatic Potential and Capacitance

JEE Main 2024 (Online) 27th January Morning ShiftNumericalMedium📈 Topic in 5% papers+4

Ques. 1JEE Main 2024, 27 January (Morning Shift)

The charge accumulated on the capacitor connected in the following circuit is (Given C = 150 ) [figure]

Answer

400

Solution

[figure]

Steady state: no current through C ⇒ each branch is a voltage divider across 10 V

VA = 10 21 + 2 = 203 V, VB = 10 46 + 4 = 4 V

VA - VB = 203 - 4 = 83 V

Q = C(VA - VB) = 150 83 = 400

Answer: 400

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Ques. 2JEE Main 2024, 27 January (Evening Shift)

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). In the light of the above statements, choose the most appropriate answer from the options given below :

Both (A) and (R) are correct and (R) is the correct explanation of (A)
(A) is correct but (R) is not correct
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(A) is not correct but (R) is correct

Ques. 3JEE Main 2024, 27 January (Evening Shift)

The electric potential at the surface of an atomic nucleus (z=50) of radius 9 10-13 cm is 106 V.

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