TBR Physics Section 8 question 23

[MrNeuro]

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A charged particle is suspended between two horizontally aligned parallel plates carrying opposite charges of equal magnitude if the separation between the plates is doubled while the voltage is held constant, what happens to the electric force on a particular drop?

A. it is doubled
B. it is halved
C. It is reduced to one fourth of its original value
D. it remains the same

Answer (highlight to see): B

so im confused the explanation uses V=Ed then F=qE

which makes no sense to me

i was wondering if you would use V=W/q= Kq/r and if that is what you use when do you use F=kqq/r^2

 

[MedPR]

A charged particle is suspended between two horizontally aligned parallel plates carrying opposite charges of equal magnitude if the separation between the plates is doubled while the voltage is held constant, what happens to the electric force on a particular drop?

A. it is doubled
B. it is halved
C. It is reduced to one fourth of its original value
D. it remains the same

Answer (highlight to see): B

so im confused the explanation uses V=Ed then F=qE

which makes no sense to me

i was wondering if you would use V=W/q= Kq/r and if that is what you use when do you use F=kqq/r^2

You only use F=Kqq/r^2 when there are two charges attracting or repelling each other. I got confused on this one too.

 

[thestrokes14]

A charged particle is suspended between two horizontally aligned parallel plates carrying opposite charges of equal magnitude if the separation between the plates is doubled while the voltage is held constant, what happens to the electric force on a particular drop?

A. it is doubled
B. it is halved
C. It is reduced to one fourth of its original value
D. it remains the same

Answer (highlight to see): B

so im confused the explanation uses V=Ed then F=qE

which makes no sense to me

i was wondering if you would use V=W/q= Kq/r and if that is what you use when do you use F=kqq/r^2

Hopefully you realized that a parallel plate capacitor is being described. Either way, there is a constant electric field between the two plates, so use equations that indicate a constant electric field (not an electric field due to point charges/Coulomb's electrostatic force).

They say that Voltage is constant, while the distance doubles. So V=Ed indicates that E and d are inversely proportional, and doubling d will half E, at constant V. We did not add any charge, so q remains the same. F=Eq indicates that E reduced by a factor of 2, with constant q, means F is reduced by a factor of 2. So the Force is cut in half.

Remember, use the right equations when you have a CONSTANT electric field.