- #1

g-racer

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For the second part I said 210=kq/r and 400=q/(4pi(r+0.1)^2) and get r=-0.1 which cannot be right! help would be appreciated thanks

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- Thread starter g-racer
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In summary, the electric potential immediately outside a charged conducting sphere can be determined by using the equations 200=kq/r and 150=kq/(r+0.1) where r is the radius of the sphere and q is the charge on it. By solving these equations, the radius is found to be 0.3m and the charge to be 6.67x10^-9C. Similarly, for another charged conducting sphere with electric potential of 210 V and electric field of 400 V/m, the equations 210=kq/r and 400=q/(4pi(r+0.1)^2) can be used to determine the radius and charge. However, there may be two solutions due to the squared

- #1

g-racer

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For the second part I said 210=kq/r and 400=q/(4pi(r+0.1)^2) and get r=-0.1 which cannot be right! help would be appreciated thanks

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- #2

Simon Bridge

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Hint: r appears squared - so you should have two solutions.

- #3

g-racer

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I know there should be two but with 1600pi r^2+(320pi-210/k)r +16pi I only get r=-0.1

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ehild

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g-racer said:For the second part I said 210=kq/r and 400=q/(4pi(r+0.1)^2) and get r=-0.1 which cannot be right! help would be appreciated thanks

The equation in red should be 400=kq(r+0.1)

ehild

- #5

Simon Bridge

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You are saying that from: 1600pi r^2+(320pi-210/k)r +16pi = 0

you only get r=-0.1 ?

But you can't...

The discriminant is (320pi-120/k)^2-4(1600pi)(16pi) < 0 ... so you should have no roots at all.

This means you set up the equation wrong.

ehild may have spotted the answer - you may have made other errors: recheck your working.

I'd start from scratch:

So you have charge q giving voltage V at radius R, and electric field E at radius R+r.

You must find q and R given V, E, and r. Leave the numbers to*after* doing the algebra.

shortcut: Write the equations in the form kq=...

you only get r=-0.1 ?

But you can't...

The discriminant is (320pi-120/k)^2-4(1600pi)(16pi) < 0 ... so you should have no roots at all.

This means you set up the equation wrong.

ehild may have spotted the answer - you may have made other errors: recheck your working.

I'd start from scratch:

So you have charge q giving voltage V at radius R, and electric field E at radius R+r.

You must find q and R given V, E, and r. Leave the numbers to

shortcut: Write the equations in the form kq=...

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- #6

g-racer

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so setting equal for kq I get 210r=400(r+0.1)^2 but then I still get negative square roots...?

- #7

Simon Bridge

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Please show your working - the trouble is either algebra or arithmetic.

Please follow advise and wait till after the algebra is done to plug the numbers in.

Start with your ##Vr=E(r+d)^2## - divide through by E, put in standard form and use the quadratic equation.

*Then* put in: V=210V, E=400V/m, d=0.1m

Please follow advise and wait till after the algebra is done to plug the numbers in.

Start with your ##Vr=E(r+d)^2## - divide through by E, put in standard form and use the quadratic equation.

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- #8

g-racer

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- #9

Simon Bridge

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This is why I kept telling you to put the numbers in last.I had put -40r for some reason

If you do all your algebra with the variables this sort of mistake is much less common and much easier to spot when it does happen.

I realize that many students are uncomfortable with this approach but it is worth it in the long run: not all your problems will be this simple.

To find the charge of a conducting sphere, you can use the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.

Yes, the charge of a conducting sphere can be negative. This indicates that the sphere has an excess of electrons, which have a negative charge.

The radius of a conducting sphere is directly proportional to its charge. This means that as the charge increases, the radius of the sphere will also increase.

The unit for charge is Coulomb (C).

Yes, the voltage of a conducting sphere can be changed by connecting it to a different voltage source. This will in turn change the charge and radius of the sphere.

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