suppose now you disconnect the electrostatic voltage source and then slowly sepa

Question

suppose now you disconnect the electrostatic voltage source and then slowly separate the plates.

i.What will happen to V, the difference in potential between the plates, and E, the magnitude of the elelctric field between the plates?
1-V increases, E remains the same
2-V remains the same, E increases
3-both V and E decrease
4-V increases, E decreases
5-V decreases, E increases
6-V decreases, E remains the same
7-V remains the same, E decreases
8-both V and E remain the same
9-both V and E increase

ii. What will happen to Q, the charge on the plates, and C, the capacitance of the system?
1-Q remains the same, C increases
2-Q increases, C decreases
3-both Q and C remain the same
4-Q increases, C remains the same
5-both Q and C decrease
6-Q remains the same, C decreases
7-both Q and C increase
8-Q decreases, C remains the same
9-Q decreases, C increases

Explanation / Answer

i) since V or the potential difference is given by V = E*d where E and d are the electric field between the plates and d is the separation of the plate. When we increase the distance the electric field magnitude do not change since from electrostatics we known that E = sigma/2Epsilon_0 which is constant for a charged conducting plate Therefore, correct option is 1-V increases, E remains the same ii) As stated before that charge of the plate will not change due to just by separation since charges have nowhere to go. Since capacitance is given by C = A*Epsilon_0/d and we are increasing d so capacitance decreases. So correct option is 6-Q remains the same, C decreases

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