# Capacitors: Series Capacitance Concepts

## Homework Statement

The sketch below is a side view of two capacitors consisting of parallel plates in air. The capacitor plates are equal in area but the plate separation differs as shown. Individual capacitors are specified with two letters, for example WX is a single capacitor. The charge on plate W is represented by QW.

(Img is http://img397.imageshack.us/img397/6074/plotkv8.png" [Broken])

E = (.5)*C*V^2
Q = C*V
C = EA/d

## The Attempt at a Solution

A potential is applied between points A and B. For each of the statements below choose the proper response.

Increase If capacitor WX is eliminated from the circuit, the the energy stored in capacitor ZY will....
It will increase because the voltage drop on ZY will increase

Increase If capacitor ZY is eliminated from the circuit, the magnitude of QX will....
Increase, voltage drop will increase

Stay the Same If the plate separation for capacitor WX increases, the energy stored in capacitor ZY will ....
The capacitance of ZY doesn't change

Less Than The potential difference across capacitor WX is ______ the potential difference across capacitor ZY.
Wx has a larger distance

Equal to QX + QZ is _____ zero.
All charges in a series capacitor are the same.

Less than The energy stored in capacitor WX is ______ the energy stored in capacitor ZY.
Because there is less voltage

Equal to The electric field between plates Z and Y is ______ the electric field between plates W and X.
Electric field is proportional to charge

I know I'm wrong. What am I doing wrong?

Last edited by a moderator:

Related Introductory Physics Homework Help News on Phys.org
Let me try again...

## Homework Statement

The sketch below is a side view of two capacitors consisting of parallel plates in air. The capacitor plates are equal in area but the plate separation differs as shown. Individual capacitors are specified with two letters, for example WX is a single capacitor. The charge on plate W is represented by QW.

(Img is http://img397.imageshack.us/img397/6074/plotkv8.png" [Broken])

E = (.5)*C*V^2
Q = C*V
C = EA/d

## The Attempt at a Solution

A potential is applied between points A and B. For each of the statements below choose the proper response.

If capacitor WX is eliminated from the circuit, the the energy stored in capacitor ZY will increase.
It will increase because the voltage drop on ZY will increase

If capacitor ZY is eliminated from the circuit, the magnitude of QX will increase.
Increase, voltage drop will increase

If the plate separation for capacitor WX increases, the energy stored in capacitor ZY will stay the same.
The capacitance of ZY doesn't change

The potential difference across capacitor WX is less than the potential difference across capacitor ZY.
Wx has a larger distance

QX + QZ is equal to zero.
All charges in a series capacitor are the same.

The energy stored in capacitor WX is less than the energy stored in capacitor ZY.
Because there is less voltage

The Electric Field between plates Z and Y is equal to the electric field between plates W and X.
Electric field is proportional to charge

I know I'm wrong.

WHAT AM I DOING WRONG?

--yuvlevental

Last edited by a moderator:
i need help

really, i do

im not joking

ill keep posting until someone answers

alphysicist
Homework Helper
Hi yuvlevental,

Here are a few problems I see:

Let me try again...

## Homework Statement

The sketch below is a side view of two capacitors consisting of parallel plates in air. The capacitor plates are equal in area but the plate separation differs as shown. Individual capacitors are specified with two letters, for example WX is a single capacitor. The charge on plate W is represented by QW.

(Img is http://img397.imageshack.us/img397/6074/plotkv8.png" [Broken])

E = (.5)*C*V^2
Q = C*V
C = EA/d

## The Attempt at a Solution

A potential is applied between points A and B. For each of the statements below choose the proper response.

If capacitor WX is eliminated from the circuit, the the energy stored in capacitor ZY will increase.
It will increase because the voltage drop on ZY will increase

If capacitor ZY is eliminated from the circuit, the magnitude of QX will increase.
Increase, voltage drop will increase

If the plate separation for capacitor WX increases, the energy stored in capacitor ZY will stay the same.
The capacitance of ZY doesn't change
The capacitance of ZY doesn't change, but the voltage across ZY depends on the capacitance of WX.

The potential difference across capacitor WX is less than the potential difference across capacitor ZY.
Wx has a larger distance
What is the relationship between plate separation and potential difference?

QX + QZ is equal to zero.
All charges in a series capacitor are the same.

The energy stored in capacitor WX is less than the energy stored in capacitor ZY.
Because there is less voltage
Why do you say there is less voltage? What is the relationship between voltage and capacitance?

The Electric Field between plates Z and Y is equal to the electric field between plates W and X.
Electric field is proportional to charge
Your reasoning gives you the right answer herem but the reasoning itself is not quite right. The electric field between the capacitors is proportional to the charge density (charge/area). For example, if in this problem one capacitor had twice the area of the other, but still the same charge, its field would be half.

Last edited by a moderator: