(a) If a spherical raindrop of radius 0.650 mm carries a charge of -1.20 pC uniformly distribute

(a) If a spherical raindrop of radius 0.650 mm carries a...

(a) If a spherical raindrop of radius 0.650 $\mathrm{mm}$ carries a charge of $-1.20 \mathrm{pC}$ uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop. (b) Two identical raindrops, each with radius and charge specified in part (a) collide and merge into one larger raindrop. What is the radius of this larger drop, and what is the potential at its surface, if its charge is uniformly distributed over its volume?

Key Concepts

Conservation of Volume and Charge in Merging Objects

When two spherical objects merge, the total charge is the algebraic sum of the individual charges, reflecting the conservation of charge. Additionally, their volumes add together, meaning the new radius can be calculated by equating the sum of the individual volumes to the volume of a new sphere. Understanding how geometric properties scale with volume is essential for analyzing changes in physical quantities, such as potential, when bodies coalesce.

Uniform Charge Distribution

Uniform charge distribution means that the electric charge is spread evenly across the volume (or the surface) of an object. This symmetry simplifies the computational analysis of the electric field and potential, as well as ensuring that the external field behaves as if all the charge were located at the center. It is a key assumption that allows for straightforward application of classical electrostatic formulas.

Electric Potential of a Charged Sphere

This concept involves finding the electrostatic potential on or outside a sphere carrying a uniform charge distribution. For a sphere, whether the charge is on the surface or uniformly distributed inside, the potential at a point outside the sphere (including on its surface) can be calculated as if all the charge were concentrated at the center, using the expression V = (1/(4???))·(Q/R). This derivation comes from applying Gauss’s law in situations with spherical symmetry.

Transcript

00:01 So in this question we have a rain job that carries a charge at its surface uniformly.

00:09 So in this ring job, it has a radius r of 0 .65 0mm, and then it carries a charge q of negative 1 .20 picol.

00:22 So it asks us to find the potential at its surface.

00:28 And we can do that by treating the charge, the uniformly distributed charge is a point charge at the center.

00:40 Why? because a sphere is a very symmetric configuration.

00:45 And for a symmetric configuration, we can, if everything's uniformly distributed, we can think of it as being in the center.

00:55 So this question part a really is asking if we have a point charge of negative 1 .2 p .cical column, what will be the potential when you are 0 .65 millimeter from it? and we've worked on the potential of point charges in section 2.

01:15 The formula we have is v equals kq over r.

01:20 So k is 9 times 10 to the netice.

01:23 Q is negative 1 .2 times 10 to the negative 12s.

01:28 R is 6 .5 times 10 to the negative force volts.

01:34 And this gives negative 16 .6 volts.

01:39 Part b, when two identical raindrops collide and merge, what happens is that you'll have a new raindrop with a different radius.

01:50 And that the radius is such that the raindrop has twice the volume of the original raindrop, because it's two raindrops merging.

01:58 So if i call this r0 and q0, v not.

02:04 So if i now have a different r, what will happen is that we want to, first let's find the relationship between a v and r.

02:14 So the volume of sphere is 3 4 pi r cubed.

02:22 So what that also means is r equals 3v over 3v over 3v over 4 pi and 2 .1 3rd.

02:37 So right now, when v becomes 2 times its original volume times 2, what this becomes is 2 to 1 3v0 over 4 pi, and this is just r.

02:58 Of not...

Please give Ace some feedback