Question

A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Do the following quantities increase, decrease, or stay the same?

Select one or more:

a. E decreases

b. Energy stored stays the same

c. C increases

d. E increases

e. Q decreases

f. delta V increases

g. Q stays the same

h. C decreases

i. Energy stored decreases

j. C stays the same

k. E stays the same

l. delta V stays the same

m. Energy stored increases

n. Q increases

o. delta V decreases

Answer 1

Answer:

g.Q remains same

f.Change in potential increases

h.C decreases.

k.E stays the same

m.Energy stored increases.

Explanation:

Let capacitance of parallel plate capacitor is C and charge stored in capacitor is Q.The distance between plate of capacitor=d

Capacitance of parallel plate capacitor=

When a parallel plate capacitor is disconnected  

Then, charge Q remains same .

$C=\frac{\epsilon A}{d}=\frac{Q}{Ed}=\frac{Q}{V}$

Where Q=Charge stored in capacitor

E=Electric field

V=Potential difference

d=Separation between plate  

Capacitance is inversely proportional to potential difference and distance between plate of capacitor.

When d increases then C decreases.

When capacitance of capacitor decreases then potential difference increases by the same factor.

$E=\frac{V}{d}$

When V increases and d decreases by the same factor  

Then, E remains constant.

Energy stored in capacitor=$\frac{1}{2}CV^2$

When C decrease and V increases then, energy stored in capacitor increases.

Answer 2

The affect of decrease in the distance between the parallel plates are; electric field increases, change in potential decreases, capacitance increases, charge stays the same, and Energy stored increases.

Charge in a capacitor

The charge in a capacitor is given by the following formulas;

$C = \frac{\varepsilon_o A}{d} = \frac{Q}{Ed} = \frac{Q}{V}$

Where;

• Q is the charge stored in capacitor
• E is the electric field
• V is the potential difference
• d is the separation between plate  

When distance, d decreases, capacitance increases.

When capacitance of capacitor increases, the potential difference decreases by the same factor.

Electric field

E = V/d

When distance, d increases, and the V decreases by the same factor, the electric field will be constant. However, at constant potential and distance decreases, the electric field will decrease.

Energy stored in capacitor

E = ¹/₂CV²

When both C and V increases the energy stored in capacitor will increase.

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