How Is the Electric Field Intensity Calculated on a Charged Sphere's Surface?

[gracy]

Homework Statement

The electric potential on the surface of a sphere of radius R and charge $3$×$10^-6$ is 500V.The intensity of electric field on the surface of the sphere (in N/C)is

Homework Equations

$V$=$(\frac{1}{4πε0}\frac{q}{R})^2$

The Attempt at a Solution

Actually I have solution but still I am unable to understand.So ,instead of my attempt at a solution .I 'll post solution itself.
$V$=$(\frac{1}{4πε0}\frac{q}{R})^2$
$E$=$\frac{\left(\frac{1}{4πε0}\frac{q}{R}\right)^2}{\frac{q}{4πε0}}$

=$\frac{25×10^4}{27×10^3}$

=$\frac{250}{27}$

I want to how the above formula of E is derived.

 

[BvU]

Where did you get these squares ? See here

 

[gracy]

Where did you get these squares ?

I didn't get you.

 

[BvU]

 

[gracy]

An image with red cross on it!

 

[BvU]

Yeah, company politics forbid easy cut & paste. I uploaded the picture from the link. Which of course you studied extensively ...

 

[gracy]

Which of course you studied extensively ...

You,king of sarcasm!

 

[gracy]

Which of course you studied extensively ...

But actually I have studied it extensively before posting this thread(hyperphysics: this site is quite popular after wiki I usually refer this for physics along with physics classroom) ,but could not figure out/extract anything useful from it as you did!

 

[gracy]

I am still confused ,please give me a hint !

 

[SammyS]

Homework Statement

The electric potential on the surface of a sphere of radius R and charge $3$×$10^-6$ is 500V.The intensity of electric field on the surface of the sphere (in N/C)is

Homework Equations

$V$=$(\frac{1}{4πε0}\frac{q}{R})^2$

Where did you get the square (as in the exponent being 2) in the above and in the following formulas?

The Attempt at a Solution

Actually I have solution but still I am unable to understand.So ,instead of my attempt at a solution .I 'll post solution itself.
$V$=$(\frac{1}{4πε0}\frac{q}{R})^2$
$E$=$\frac{\left(\frac{1}{4πε0}\frac{q}{R}\right)^2}{\frac{q}{4πε0}}$

=$\frac{25×10^4}{27×10^3}$

=$\frac{250}{27}$

I want to how the above formula of E is derived.

 

[gracy]

Where did you get the square (as in the exponent being 2) in the above and in the following formulas?

It has been given in the solution.

 

[BvU]

I don't see no solution.

 

[ehild]

Homework Statement

The electric potential on the surface of a sphere of radius R and charge $3$×$10^-6$ is 500V.The intensity of electric field on the surface of the sphere (in N/C)is

Homework Equations

$V$=$(\frac{1}{4πε0}\frac{q}{R})^2$

The Attempt at a Solution

Actually I have solution but still I am unable to understand.So ,instead of my attempt at a solution .I 'll post solution itself.
$V$=$(\frac{1}{4πε0}\frac{q}{R})^2$
.

Gracy, the equation for the potential is wrong. You either copied it incorrectly or your book is wrong. You should know the potential formula for a charged sphere.

 

[gracy]

You either copied it incorrectly or your book is wrong.

I copied it incorrectly.I know $V$=$\frac{1}{4πε0r}$ It was just by mistake.But I want to know about formula of E

 

[BvU]

And what is it you want to know ?

 

[gracy]

I want to how the above formula of E is derived.

 

[gracy]

I want to make sure there is square in the formula of E!

 

[BvU]

Sure. But by squaring $q\over 4\pi\epsilon_0$ you (or the 'solution writer') grabbed the wrong factor !
Check the dimensions too -- very helpful !

 

[gracy]

I don't know what is denominator in the formula of E?

 

[gracy]

numerator is $V^2$?

 

[gracy]

I don't know what is denominator in the formula of E?

The intensity of electric field on the surface of the sphere (in N/C)is

According to his denominator is charge but how can this $\frac{1}{4πε0}$ be charge?

 

[BvU]

Electric field strength is Volt/meter, a.k.a. Newton/Coulomb

 

[gracy]

But I am told to express E in Newton/Coulomb

 

[BvU]

Yes. One and the same thing. Volts is Joule/Coulomb, Electric field is Newton/Coulomb and sure enough, Joule/meter = Newton !

 

[gracy]

Here denominator is coulomb means in formula there should be charge as denominator,right?

 

[BvU]

No. You also don't have a charge in the denominator of V and that is Joules/Coulomb !

Ever unraveled the dimension of $\epsilon_0$ ?

A very good place is this

 

[gracy]

I know unit of $ε0$
$C^2$/$N$.$m^2$

 

[BvU]

Right. So does the dimension of V match your expectations ? And then: what's the difference with the dimension of E ?

 

[gracy]

Dimension for Voltage is $ML^2T^3 I^-1$ Dimension of E $MLT^3I^-1$

what's the difference

Difference of one L (length).

 

[gracy]

I am still clueless about my question in op.
$E$=$\frac{\left(\frac{1}{4πε0}\frac{q}{R}\right)^2}{\frac{q}{4πε0}}$

I want to how the above formula of E is derived.

 

[BvU]

And when we flash back to post #4 ...

 

[BvU]

That's flashing back even further !

But the same link has something to say about the derivation too, I seem to remember. The name Gauss comes to mind !
I take it you can look around there all on your own (with a little patience, perhaps ...) , so -- like a good helpdesk worker -- I can now safely close this ticket, I hope ?

 

[gracy]

Do you remember Coulomb's Law?

$F$=$\frac{K Qq}{r^2}$

 

[haruspex]

I am still clueless about my question in op.
$E$=$\frac{\left(\frac{1}{4πε0}\frac{q}{R}\right)^2}{\frac{q}{4πε0}}$

They took the equation $E$=$\frac{1}{4πε_0}\frac{q}{R^2}$ and rewrote it in that form. You can check they are equivalent. This was useful because the numerator is the same as V² and the denominator only involves the charge and some constants. Since V and q are known, this gives a quick way of finding E.