COMEDK · Physics · 18. Capacitance

A capacitor of capacitance 8\(\mu\)F is fully charged by connecting it to a source of 200V. It is then disconnected from the supply and connected to an uncharged capacitor of capacitance 4\(\mu\)F. The electrostatic energy lost in this sharing is:

A \(10.67 \times 10^{-2}\ J\)
B \(5.33 \times 10^{-2}\ J\)
C \(3.53 \times 10^{-3}\ J\)
D \(21.34 \times 10^{-2}\ J\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(5.33 \times 10^{-2}\ J\)

Step-by-step Solution

Detailed explanation

Given: \(C_1 = 8 \mu F = 8 \times 10^{-6}\ F\) \(V_1 = 200\ V\) \(C_2 = 4 \mu F = 4 \times 10^{-6}\ F\) \(V_2 = 0\ V\) The loss of electrostatic energy when two capacitors are connected is given by: \(\Delta U = \dfrac{1}{2} \dfrac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From COMEDK

Explore more questions on app