A dielectric slab is slowly inserted between the plates of a parallel plate capa

Question

A dielectric slab is slowly inserted between the plates of a parallel plate capacitor, while the potential difference between the plates is held constant by a battery. As it is being inserted: a. the capacitance, the potential difference between the plates, and the charge on the positive plate all increase. b. the capacitance, the potential difference between the plates, and the charge on the positive plate all decrease. c. the potential difference between the plates increases, the charge on the positive plate decreases, and the capacitance remains the same. d. the capacitance and the charge on the positive plate decrease but the potential difference between the plates remains the same. e. the capacitance and the charge on the positive plate increase but the potential difference between the plates remains the same A capacitor with capacitance C will charge most quickly when put in series with a resistance of a. 200 k ohm b. 100 k k ohm c. 50 k ohm d. 10 k ohm e. 150 k ohm

Explanation / Answer

whenever a dielectric is inserted in a capacitor, it's capacitance increases.

the potential difference is given to be constant in the question

we know that Q = CV

from here we deduce that Q increases as C increases

so (E) will be the correct option

6.) the time taken to charge by 63% is called time constant T = RC

if R is small, time taken will be small and thus it'll charge fastly.

so, (D) 10kohm should be the resistance

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