A parallel plate capacitor is charged and the charging battery is then...
Yes ans is B and D because once a capacitor is charged and after disconnecting the battery,the charge flowing in the circuit will be constant.
also C is inversely proportional to d. so as d is increased C gets decreased and voltage =Q/C. so as Q remains constant so voltage is inversely proportional to C. so as C is decreasing on increasing distance between the plates voltage will get increased.
and energy =Q^2/2C so energy is also inversely proportional to C so it will also get increased on increasing distance.
A parallel plate capacitor is charged and the charging battery is then...
Explanation:
Introduction:
A parallel plate capacitor consists of two conducting plates separated by a dielectric material. When a capacitor is charged, it stores electrical energy in the form of electric field between its plates. In this scenario, the plates of the capacitor are moved farther apart after charging and the charging battery is disconnected.
Bullet Points:
To understand the given options, let's analyze each one:
a) The charge on the capacitor increases:
Moving the plates farther apart does not cause any change in the charge stored in the capacitor. Charge is conserved and remains constant unless some external influence is applied. Therefore, this option is incorrect.
b) The voltage across the plates increases:
Moving the plates farther apart increases the distance between them. As a result, the electric field between the plates decreases. Since voltage is directly proportional to electric field, the voltage across the plates increases. Therefore, this option is correct.
c) The capacitance increases:
The capacitance of a parallel plate capacitor is given by the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. In this scenario, the area of the plates and the permittivity of free space remain constant. As the plates are moved farther apart, the distance between them (d) increases. Therefore, the capacitance decreases according to the formula. Thus, this option is incorrect.
d) The electrostatic energy stored in the capacitor increases:
The energy stored in a capacitor is given by the formula E = (1/2)CV², where C is the capacitance and V is the voltage across the plates. As mentioned earlier, the voltage (V) increases when the plates are moved farther apart. Since the capacitance (C) remains constant in this scenario, the electrostatic energy (E) stored in the capacitor increases. Therefore, this option is correct.
Conclusion:
Based on the analysis, options b) and d) are correct. Moving the plates farther apart increases the voltage across the plates and consequently increases the electrostatic energy stored in the capacitor.
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