InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What will be the effect on the potential between the plates of a charged parallel plate capacitor when both plates are carry nearest to each other. While the charge is constant. Explain? |
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Answer» q = CV |
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| 2. |
The potential energy between plates of a charged parallel plate capacitor is UQ. If a dielectric plate of dielectric constant er is placed between The gap of plates, then the new potential energy will be :(a) \(\frac{U_{0}}{\varepsilon_{r}}\)(b) U0εr2(c) \(\frac{U_{0}}{\varepsilon_{r}^{2}}\)(d) U0 |
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Answer» (a) \(\frac{U_{0}}{\varepsilon_{r}}\) |
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| 3. |
What is the resultant electric field between the plates of a charged parallel plate capacitor when the surface charge density of plates is σ? |
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Answer» Resultant electric field E = \(\frac{\sigma}{\varepsilon_{0}}\) |
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| 4. |
On what factors does the capacitance of a conductor depend? |
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Answer» The capacitance of a conductor depend on area of cross-section and medium surrounding the conductor. |
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| 5. |
Eight Mercury drops of equal radius and equal charge combine to form a big drop. The capacitance of big drop in comparison the each small drop will be:(a) 2 times(b) 8 times(c) 4 times(d) 16 times |
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Answer» (a) 2 times |
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