Question

The electric potential is $40V$at point A near a uniformly charged sphere. At point B, $2.0\mu m$farther away from the sphere, the potential has decreased by 0.16 mV. How far is point A from the center of the sphere?

Short answer

Point A is 49cm far from center of the sphere.

Step-by-step solution

Given Information 

We have given that the electric potential is 40 V at point A near a uniformly charged sphere. At point B, $20\mu m$farther away from the sphere, the potential has decreased by 0.16 mV.

Simplify

The potential at point A, $V_{a,}$and the potential at point B, which is $△r$away from point A, is $△V$lower and the distance of point A from the charge creating the potential.

The definition of the potential applied for point A, we can get the following:

$V_a=\frac{kq}{r_a}\Rightarrow r_a=\frac{kq}{V_a},$

where $r_a$is what we want to find.

The difference in potential, we can obtain the following:

$△V=V_a-V_b=kq\left(\frac{1}{r_a}-\frac{1}{r_a+△r}\right)=\frac{kq△r}{r_a(r_a+△r)}$

The product of the two distance,

$r_a\left(r_a+△r\right)=\frac{kq△r}{△V}$

The charge q in unknown here, However the first expression by substituting it at the first factor we'll get

$\frac{kq}{V_a}\left(r_a+△r\right)=\frac{kq△r}{△V}$

This allows to reach the conclusion that

$r_a+△r-△r\frac{V_a}{△V}$

Passing the difference in position to the right gives us the final expression:

$r_a=△r\frac{V_a-1}{△V}$

We can assure ourselves our solution is not wrong, since the units of the two sides are the same: on the right side, both potentials cancel, while the length remains on both sides.

Given numerical solution

$r_a=2.{10}^{-6}·\frac{40-1}{1.6·{10}^{-4}}=49cm$