(a) If a spherical raindrop of radius 0.650 mm carries a charge of -3.60 pC uniformly distribute

step 1 Welcome to problem number 23 .72. So we have a spherical raindrop of radius 0 .650 millimeter ha

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

Key Concepts

Electric Potential of a Uniformly Charged Sphere

This concept refers to the method of calculating the electric potential at the surface of a sphere that has its charge distributed uniformly throughout its volume. In the case of points outside such a sphere, the sphere behaves as if all of its charge were concentrated at its center, allowing one to calculate the potential using the formula for a point charge. This concept is crucial for understanding how the spatial distribution of charge affects the potential, particularly when comparing situations involving uniform versus surface charge distributions.

Gauss’s Law

Gauss’s Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It simplifies the calculation of electric fields and potentials for systems with a high degree of symmetry, such as spherical charge distributions. This law provides the theoretical basis for treating a uniformly charged sphere as if its total charge were located at its center when calculating the external potential.

Conservation of Charge

The principle of conservation of charge states that the total electric charge in an isolated system remains constant over time, regardless of any changes that occur within the system, such as the merging or splitting of charged objects. When two charged bodies combine, the total charge of the resulting body is simply the sum of the individual charges. This concept is key when analyzing processes like the collision and merging of raindrops or other charged spheres.

Volume Scaling in Spherical Merging

When two or more spherical objects merge, the total volume of the resulting sphere is the sum of the individual volumes. Since the volume of a sphere is proportional to the cube of its radius, any change in volume leads to a change in radius according to the cube root of the overall volume ratio. This concept is important for determining the new radius after merging, given that the charge is uniformly distributed throughout the volume.

Transcript

00:01 Welcome to problem number 23 .72.

00:04 So we have a spherical raindrop of radius 0 .650 millimeter having charge equal to minus 3 .60 picoculum.

00:15 So we are asked to find out the value of the potential at the surface of this reendrop.

00:21 So this is raindrop.

00:24 This is r or capital r having charge q.

00:30 So we are asked to find out the potential on its surface.

00:33 So let's say we have point here and we have this formula here which is equal to 1 by 4 pi a epsilon q over capital r so we have q distributed of the whole volume of the spherical shape capital r is the radius of this spherical shape the potential on the surface equals to 1 by 4 by upside not capital k capital k so we can use this formula to find out the value of the potential so we have potential on the surface of the rainbow equals to 9 into 10 x to power 9 now putting the value the q here so which is minus 3 .60 into 10 rest to power minus 12 1 pko is equal to 10 to minus 12 so we have this value here over r which is 0 .650 millimeter and if we want to convert into the si units so this becomes 0 .65 into 10 rest of minus 3 meter so this becomes 0 .65 into 10 to minus 3 meter.

01:54 So using the calculator, i have found the value.

01:57 So this is equals to minus 49 .8 volt.

02:09 So that's your answer for part a.

02:11 So there was only single raindrop here.

02:14 So single raindrop or single sphere.

02:23 Now the part b of this question says that if two identical raindrops, each with the radius and charts, specified in part a so with the same radius in q what is the radius of this larger drop so let's do part b now so this was part a so the part b says if two raindrop add and makes a bigger raindrop that means the volume of this sphere sphere one and the volume of this sphere so let's say with sphere 2 must be equal to the volume of the bigger sphere so we have 4x3 pi r cube plus 4 by 3 pi r cube because both have r radius now let's say we have r prime so this is our new radius r prime this has volume equals 4 pi 3 pi r prime cube so this gives us 4 pi r cube goes away so this gives us 2 r cube equals to r prime cube this gives us r prime equals to two power one over three r so that's your new radius now this has q charge and this has q charge so total charge is q plus q which is equal to 2 q so the charge is conserved so we have new charge let's say we have q prime which is equal to 2 q or say q new so we have 2 times minus 3 .6 into 10 rest of minus 12 column.

04:44 So which becomes minus 7 .2 into 10 rest of minus 12 column.

04:51 So that's your q prime.

04:53 That's your new charge on this sphere.

04:56 So this carries q prime charge and radius r prime.

05:01 All right.

05:02 Now we are asked to find out the value of the potential again on the surface.

05:05 So we have this formula here again using the same formula.

05:09 But q and r changes...