# Test tommorow : electic charge

• XxXDanTheManXxX
I think he needs to go back and learn how to derive everything. In summary, the conversation discusses a problem involving an isolated spherical conductor of radius 20cm charged to 4k V. The conversation provides equations for calculating the charge and capacitance of the sphere, as well as how the capacitance changes if the sphere is charged to 8k V. There is also a suggestion to review how to derive equations instead of simply memorizing them.
XxXDanTheManXxX
Studying help

An isolated spherical conductor of radius 20cm is charged to 4k V (i.e. 4000 V).
a. How much charge is on the conductor?
b. What is the capacitance of the sphere?
c. how does the capacitance change if the spere is charged to 8 k V (8000)?

b) Use the relationship Q = CV to find C.
c) Use the solved equation above to find the effect on C if V = 2V

Since you have test tomorrow I do not know if this will be of any help, but here it is anyway.
a=R=20cm

a.)
$$4000 V=- \int_\infty^a \vec{E} dr =-\int_\infty^a \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} dr$$

b.)
$$Q= C \Delta V$$
$$\frac{Q}{\Delta V}= C$$
$$\Delta V =-\int_\infty^a \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} dr$$
steps not shown
$$C=4 \pi \epsilon_0 R$$

3.) What Whozum did not mention here is that Q and V are related so it is not as easy as he says it is. I provided about an equation that should work just fine for you.

Davorak:

With $$V = \frac{Q}{4\pi\epsilon_0 R} \mbox{ and } C = \frac{Q}{V} = 4\pi\epsilon_0R$$

$$2V = \frac{Q}{2\pi\epsilon_0 R} \mbox{ so then } C = \frac{Q}{2V} = 2\pi\epsilon_0R$$ ?

The charges cancel in either case, no?

whozum said:
Davorak:

With $$V = \frac{Q}{4\pi\epsilon_0 R} \mbox{ and } C = \frac{Q}{V} = 4\pi\epsilon_0R$$

$$2V = \frac{Q}{2\pi\epsilon_0 R} \mbox{ so then } C = \frac{Q}{2V} = 2\pi\epsilon_0R$$ ?

The charges cancel in either case, no?
You are considering Charge and Voltage independant of each other when they are not.

$$2V = \frac{2 Q}{4\pi\epsilon_0 R} \mbox{ so then } C = \frac{2 Q}{2V} = 4\pi\epsilon_0 R$$

When you increase Q you increase V and vs versa for this problem.

So doubling the voltage has no effect on capacitance because doubling the voltage doubles the charge?

By George I think you’ve got it. Really vs versa though since charge is usually considered fundamental while voltage is not.

XxXDanTheManXxX said:
Studying help

An isolated spherical conductor of radius 20cm is charged to 4k V (i.e. 4000 V).
a. How much charge is on the conductor?
b. What is the capacitance of the sphere?
c. how does the capacitance change if the spere is charged to 8 k V (8000)?

You appear to have posted a SERIES of homework questions on here - many of them actually are related, meaning that if you understand how to solve one, the others should follow along the same line. However, in all of the questions you have posted, not ONCE have you even described your attempt or what you have tried in solving the problem, nor are we told what you do know. I mean, you MUST have some clue on where to start, or else we would just be doing the home work for you and doing you a complete disservice.

Please READ the STICKY for this section of PF before you post more questions.

Zz.

Not to disagree with ZapperZ, nevertheless this is all the OP should need

$$V=k_eq/r$$

GeneralChemTutor said:
Not to disagree with ZapperZ, nevertheless this is all the OP should need

$$V=k_eq/r$$

And not to disagree with you either, but if the OP does not even know this, then "Houston, we have a problem!"

Zz.

ZapperZ said:
GeneralChemTutor said:
Not to disagree with ZapperZ, nevertheless this is all the OP should need
$$V=k_eq/r$$

And not to disagree with you either, but if the OP does not even know this, then "Houston, we have a problem!"

Zz.
Knowing that would certainly help him for spherical cases, but he does not seem to know how to do it general. On another thread he was unaware how to get a electric field in a parallel plate capacitor. This is why I gave the more general form. He seems to be memorizing equations rather knowing how to derive them.

Last edited:

## 1. What is electric charge?

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. It can be positive or negative and is measured in units of coulombs (C).

## 2. How is electric charge measured?

Electric charge is measured using an instrument called an electrometer. The amount of charge can be calculated by measuring the force between two charged objects and using Coulomb's law.

## 3. What is the difference between static and current electricity?

Static electricity is the build-up of electric charge on the surface of an object, while current electricity is the flow of electric charge through a conductor. Current electricity is caused by the movement of electrons, while static electricity is caused by an imbalance of positive and negative charges.

## 4. How is electric charge transferred?

Electric charge can be transferred through direct contact, induction, or friction. In direct contact, the charges move from one object to another. In induction, the charges are rearranged within an object due to the presence of an external electric field. Friction occurs when two objects rub against each other, causing a transfer of electrons.

## 5. What are the effects of electric charge?

Electric charge has several effects, including the attraction or repulsion between charged objects, the creation of electric fields, and the production of electric current. It also plays a crucial role in many everyday phenomena, such as lightning, static cling, and chemical reactions.

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