Is the Potential Zero Inside a Grounded Inner Sphere?

[loui1410]

Hello,

I have my first exam tomorrow in Electricity & Magnetism and I really need to know whether I solved this question correct or not, and I do not own the answer key to the book which I took the question from. Thanks in advance!

Note: English is not my mother-tongue and neither do I study physics in English. That is, I have translated the following question. Excuse me for any errors in the use of scientific terms.

-------------

A spherical conductor, whose radius is 10cm, carries a charge of +8*10^-8 C, and another spherical conductor, whose radius is 6cm, is placed inside the first one and shares with it the same centre. The inner sphere is grounded.
1- Calculate the charge that is on the inner sphere.
2- Calculate the potential that is on surface of the outer sphere.
3- Calculate the potential and the strength of the electric field at 4cm distance from the centre of the two spheres.
4- Calculate the potential and the strength of the electric field at a point distanced 20cm from the centre of the two spheres.
5- Calculate the required work in order to transfer a charge of -5*10^-6 from the surface of the outer sphere to the surface of the inner sphere.

-------------

I have studied the required material in a really short period of time, so I am a bit confused at the moment and might ask really stupid questions, and my following attempts for solving the question may be too dumb :P

1- I there is no charge on the inner sphere because it is grounded.

2- V = k*q/r
V = 9*10^9 * (8*10^-8)/(0.1)
V = 7200 V

Is that correct? :S

3- I am confused about this one. If my answer to section 1 was correct, aren't the potential and the strength of the electric field supposed to be zero as well?

4- V = k*q/r
V = 9*10^9 * (8*10^-8)/(0.2)
V = 3600 V

E = V/r = 18000 V/m

5- W = q(V-V') = -5*10^-6 * (7200-0) = -0.036 J

 

[anathema]

Hello,

Thank you for reaching out and sharing your questions. I will do my best to help you understand the concepts and solve the problems correctly.

1- Your answer is correct. Since the inner sphere is grounded, it will have no net charge on it.

2- Your calculation is correct. The potential at the surface of the outer sphere can be calculated using the equation V = k*q/r, where k is the Coulomb's constant, q is the charge on the sphere, and r is the radius of the sphere.

3- The potential and electric field at a point between the two spheres will not be zero. This is because the outer sphere still has a net charge on it, which will create an electric field around it. The potential and electric field at this point can be calculated using the superposition principle, which states that the total potential and electric field at a point is the sum of the individual contributions from each charge. So, you would need to calculate the potential and electric field due to the charge on the outer sphere and the charge on the inner sphere separately, and then add them together.

4- Your calculations are correct for this part as well.

5- Your calculation is correct for this part too. The work done in transferring a charge from one point to another can be calculated using the equation W = q(V-V'), where q is the charge, V is the potential at the initial point, and V' is the potential at the final point.

I hope this helps you understand the concepts better and feel more confident for your exam tomorrow. Good luck!

 

[baba89]

Hello,

Thank you for reaching out for help with your exam preparation. I am a scientist and I would be happy to assist you with your questions in electrostatics.

1- Your answer is correct. Since the inner sphere is grounded, it has no net charge on it.

2- Your calculation is correct. The potential on the surface of the outer sphere is 7200 V.

3- The potential and the strength of the electric field at 4cm distance from the centre of the two spheres will not be zero. This is because the outer sphere still has a charge on it and it will create an electric field. The potential can be calculated using the same formula as before (V=k*q/r) and the electric field can be calculated using the formula E=V/r.

4- Your calculations for the potential and electric field at a distance of 20cm from the centre of the two spheres are correct.

5- Your calculation for the required work is also correct. In order to transfer a charge of -5*10^-6 C from the surface of the outer sphere to the surface of the inner sphere, a potential difference of 7200 V is required and the work done is -0.036 J.

I hope this helps clarify your doubts. Remember to double check your calculations and units to ensure accuracy. Best of luck on your exam!