If we placed a dielectric into a capacitor while it is connected to the battery

Question

If we placed a dielectric into a capacitor while it is connected to the battery and increased the distance of plate separation by 2, determine if the following increase, decrease, stay the same, or cannot be determined by the information given when comparing the values before the dielectric is added to after the battery is disconnected.

Quantity

Increase

Stay the same

Decrease

Cannot be determined

Voltage between the plates

X

Total electric field strength between the plates

X

Magnitude of charge on the plates

X

Electric field strength between the plates from the plates

X

The electric potential of the negatively charged plate

X

A-E are already categorized, I need an explanation as to why that is the correct answer.

Quantity

Increase

Stay the same

Decrease

Cannot be determined

Voltage between the plates

X

Total electric field strength between the plates

X

Magnitude of charge on the plates

X

Electric field strength between the plates from the plates

X

The electric potential of the negatively charged plate

X

Explanation / Answer

a)

Voltage between the plates stays the same since battery remains connected so voltage remains constant , while seperation between plates increased by a factor 2

b)

electric field is given by

E=V/d

since voltage is constant ,electric field is inversly proportional to seperation between the plates .therefore increasing seperation between the plates by a factor of 2 will result in decrease in electric field by a factor of 2..

c)

capacitance is given by

C=eoA/d

capacitance is inversly proportional to seperation between the plates ,so capacitance decreases by a factor of 2.

since charge Q=CV

charge is directly proportional to capacitance ,therefore it decreases by a factor of 2.

d)

Electric field strength between the plates from the plates-cannot be determined

e)

The electric potential of the negatively charged plate stays the same

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