Embibe Experts Solutions for Chapter: Capacitance, Exercise 1: Exercise-1

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Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Capacitance, Exercise 1: Exercise-1

Attempt the practice questions on Chapter 23: Capacitance, Exercise 1: Exercise-1 with hints and solutions to strengthen your understanding. Alpha Question Bank for Medical: Physics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Capacitance, Exercise 1: Exercise-1 with Hints & Solutions

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Two parallel plates capacitors of value C and 2C are connected in parallel and are charged to a potential difference V. If now battery is disconnected and a medium of dielectric constant K is introduced between the plates of the capacitor C, then the potential difference between the capacitors plates will become-

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As shown in the figure half the space between plates of a capacitor is filled with an insulator material of dielectric constant K, if initial capacity was C then the new capacity is, 

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Two materials of dielectric constant k1 and k2 are filled between two parallel plates of a capacitor as shown in the figure. The capacity of the capacitor is:

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After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates

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A parallel plate capacitor is filled with two dielectrics as shown in figure. If A is area of each plate, then the effective capacitance between X and Y is

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An insulator plate is passed between the plates of a capacitor. Then current

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In the adjoining diagram, two geometrically identical capacitors A and B are connected to a battery. Air is filled between the plates of C1 and a dielectric is filled between the plates of C2, then

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The electric field between two parallel plates of a capacitor is 2.1×10-5NC-1. If a medium is inserted between the plates than the electric field becomes 10×10-5NC-1. Now, the value of dielectric will be