AP EAMCET · PHYSICS · Capacitance
A capacitor has capacitance \(C_0\) when there is no dielectric between it's plates. 2 slabs of dielectric constant \(K_1, K_2\) respectively with area equal to area of plates but thickness half of the distance between the plates are placed in between the plates. Then the new capacitance is
- A \(C_0\left(K_1+K_2\right)\)
- B \(C_0\left(\frac{K_1 K_2}{K_1+K_2}\right)\)
- C \(C_0\left(\frac{K_1+K_2}{K_1 K_2}\right)\)
- D \(2 C_0\left(\frac{K_1 K_2}{K_1+K_2}\right)\)
Answer & Solution
Correct Answer
(D) \(2 C_0\left(\frac{K_1 K_2}{K_1+K_2}\right)\)
Step-by-step Solution
Detailed explanation
Initial capacitance of capacitor, \(C_0=\frac{\varepsilon_0 A}{d}\) Now, capacitor is filled with two dielectrics each slab has area \(A\) and distance \(d / 2\) as shown Above arrangement is a series combination of two capacitors as shown so, capacity of above arrangement is…
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