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Radius of a charged sphere

Can anyone help me with this please Ive tried using the equation r = root Q/4piε E to no avail
Two metal spheres are almost touching and have charges of +4.0 nC and -5.2nC. Given that the radius of one of the spheres is 1.8 cm and the force between them is 1x10-4 N, calculate the radius of the second sphere.
use F=q1q24πϵ0r2 F = \dfrac{q_1 q_2}{4\pi \epsilon_0 r^2}
so rearranging gives r = root q1q2/4xpixεxF ? because when i plug the values in i get math error on the calculator :confused:
Original post by mathsisfun?
so rearranging gives r = root q1q2/4xpixεxF ? because when i plug the values in i get math error on the calculator :confused:

what is the distance between the centres of the 2 spheres?
replace then solve for the radius of the second sphere
Are you getting a maths error as you are putting charges as + and -?
As an attractive force, it should actually be - 1.0 x 10-4 N.
So make sure you are square rooting a positive number!
Still stuck on this one where does the 1.8cm come into ive tried the above and now with a negative which gives me a huge number but im guessing i need to do something involving the given radius?
You are trying to calculate the separation (r) of centres. You then subtract one sphere's radius to find the other's radius (as they are touching).
Post your working? (You spotted nano I assume, and have a value for epsilon which is small!)
Original post by mathsisfun?
Still stuck on this one where does the 1.8cm come into ive tried the above and now with a negative which gives me a huge number but im guessing i need to do something involving the given radius?

Use the magnitude of the charges, not the sign or you will square root a negative number

F=q1q24πϵr2 F = \dfrac{q_1 q_2}{4 \pi \epsilon r^2}
r represents the distance between the centres of the 2 spheres
thus r = (1.8 + x) x10^-2 m

so (1.8+x)×102=(q1q24πϵ0F)0.5 (1.8 + x) \times 10^{-2} = (\dfrac{q_1 q_2}{4 \pi \epsilon_0 F})^{0.5}
solve for x, the radius of the second sphere
x = 0.025 m = 2.5 cm
(edited 4 years ago)
Original post by mathsisfun?
Can anyone help me with this please Ive tried using the equation r = root Q/4piε E to no avail
Two metal spheres are almost touching and have charges of +4.0 nC and -5.2nC. Given that the radius of one of the spheres is 1.8 cm and the force between them is 1x10-4 N, calculate the radius of the second sphere.


Original post by BobbJo
Use the magnitude of the charges, not the sign or you will square root a negative number

F=q1q24πϵr2 F = \dfrac{q_1 q_2}{4 \pi \epsilon r^2}
r represents the distance between the centres of the 2 spheres
thus r = (1.8 + x) x10^-2 m

so (1.8+x)×102=(q1q24πϵ0F)0.5 (1.8 + x) \times 10^{-2} = (\dfrac{q_1 q_2}{4 \pi \epsilon_0 F})^{0.5}
solve for x, the radius of the second sphere
x = 0.025 m = 2.5 cm


Original post by old_teach
You are trying to calculate the separation (r) of centres. You then subtract one sphere's radius to find the other's radius (as they are touching).
Post your working? (You spotted nano I assume, and have a value for epsilon which is small!)



I was shocked about the answer/comment given by the people who may not (or don’t) even give a good thought about the question.

If the question just asked the students to find the radius of the second sphere only, the author of the question and the people who just answer the question (without questioning the question) do not understand physics or maybe just electrostatics. Why? Because the spheres are conducting and the separation between the two spheres is comparable to the dimension of the sphere. These imply that we cannot apply Coulomb’s law. Most importantly, Coulomb’s law defines the force of electrostatic interaction between two point charges.

If the question has a second part in getting the students to think about the validity of the answer, then I would agree that students can solve using Coulomb’s law.

Or else (IMO) it is a faulty question at A level but it is not faulty at the university level because university physics students can solve by imposing some boundary conditions. Again this is beyond A level.
Thank you! been stuck on this one for ages even my maths teacher couldn't get it :gthumb:

(Original post by BobbJo)
Original post by BobbJo
Use the magnitude of the charges, not the sign or you will square root a negative number

F=q1q24πϵr2 F = \dfrac{q_1 q_2}{4 \pi \epsilon r^2}
r represents the distance between the centres of the 2 spheres
thus r = (1.8 + x) x10^-2 m

so (1.8+x)×102=(q1q24πϵ0F)0.5 (1.8 + x) \times 10^{-2} = (\dfrac{q_1 q_2}{4 \pi \epsilon_0 F})^{0.5}
solve for x, the radius of the second sphere
x = 0.025 m = 2.5 cm

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