Two parallel plate capacitors are connected in parallel with each other and with

Question

Two parallel plate capacitors are connected in parallel with each other and with a battery of voltage V. The first capacitor (C1) has area A and separation distance d. The second (C2) has twice the area of C1, but only half the separation distance.

a.) If the electric field strength between the plates in C1 is E, then that between the plates in C2 is:

E/4, E, E/2, 2E, 4E

b.) If the capacitance of C1 is C then that of C2 is:

C4, C, C/4, 2C, C/2

c.) If the total charge on the top plate of C1 is +Q, that on the top plate of C2 is:

Q4, Q, Q/4, 2Q, Q/2

Explanation / Answer

A)

Potential difference across the capacitors is same that is V.

the electric filed between the plates is

E=V/separation between the plates

Therefore, the electric filed inside the second capacitor is double that of first capacitor that is

E2=2E1.

B)

Capacitance of a parallel plate capacitor

C=e0A/d

So, C2=eo(2A)/(d/2)=4C1

C)

E1=Q1/e0A. And E2=Q2/e0(2A)

But E2=2E1

2Q1/e0A.=Q2/e0(2A)

Q2=4Q1=4Q

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