Calculating electrical potential difference with a sphere

[HenryHH]

Homework Statement

A sphere with radius 2.0 mm carries a 1.0 μC charge. What is the potential difference, VB - VA, between point B 4.0 m from the center of the sphere and point A 9.0 m from the center of the sphere? (The value of k is 9.0 × 10^9 N∙m2/C2.)

Homework Equations

The formula for electrical potential difference: V = (ke x Q)/r

The Attempt at a Solution

My professor said the answer is 1300V, but I'm having a hard time arriving at that answer. My understanding is that I should use the formula V = (ke x Q)/r first for point B and then for point A, and then I should subtract the answer I got for point A from the answer I got for point B. However, I'm not sure what to plug in for the "r" value of the formula. After converting from mm to m, the radius of the sphere is .002m. If point B is 4m from the center of the sphere and point A is 9m from the center of the sphere, do I subtract .002 from each of those numbers? Unless I have just incorrectly converted units, I am not getting an answer anywhere near 1300V. Is there some other formula I need to use?

 

[ehild]

My professor said the answer is 1300V, but I'm having a hard time arriving at that answer. My understanding is that I should use the formula V = (ke x Q)/r first for point B and then for point A, and then I should subtract the answer I got for point A from the answer I got for point B. However, I'm not sure what to plug in for the "r" value of the formula.

The charged sphere can be taken as if its whole charge were concentrated in the centre. So you have to plug in the given distances from the centre, 4 m and 9 m.

ehild

 

[HenryHH]

Thanks. So why was the radius of the sphere given? Just to throw off the fact that the problem is actually easier and more straightforward than it looks?

 

[ehild]

It is kind of knowledge that the radius of the sphere is irrelevant when you need to find the electric field outside the sphere.

The charged sphere can be substituted with a point charge in the centre when the charge distribution has spherical symmetry. Not otherwise.

Sometimes a problem asks the electric field or potential inside a charged sphere. In that case you have to take the charge distribution and the radius into account.

ehild

 

[asteriatic]

Good question! The formula V = (ke x Q)/r is correct, but the value of "r" that you should use is the distance from the center of the sphere to the point where you want to calculate the potential difference. In this case, for point B it would be 4.0 m and for point A it would be 9.0 m. You do not need to subtract the radius of the sphere (.002 m) from these values.

So, for point B, the potential difference would be V = (9.0 x 10^9 N∙m^2/C^2 x 1.0 x 10^-6 C)/4.0 m = 2.25 x 10^3 V.

Similarly, for point A, the potential difference would be V = (9.0 x 10^9 N∙m^2/C^2 x 1.0 x 10^-6 C)/9.0 m = 1.0 x 10^3 V.

Therefore, the potential difference between point B and point A would be VB - VA = 2.25 x 10^3 V - 1.0 x 10^3 V = 1.25 x 10^3 V, which is equivalent to 1250 V. This is different from the answer your professor gave, 1300V, but it could be due to rounding errors or using a slightly different value for the constant k.

Overall, your approach was correct, but just remember to use the distance from the center of the sphere to the point of interest, rather than subtracting the radius of the sphere from the distance. Keep up the good work!