Two metal spheres each 1.0 cm in radius are far apart. One sphere carries 38 nC, the other -10 n

step 1 Let's start with the explanation of this particular problem. We'll try to jot down the data firs

Two metal spheres each 1.0 cm in radius are far apart. One...

Two metal spheres each $1.0 \mathrm{cm}$ in radius are far apart. One sphere carries 38 nC, the other -10 nC. (a) What's the potential on each? (b) If the spheres are connected by a thin wire, what will be the potential on each once equilibrium is reached? (c) How much charge moves between the spheres in order to achieve equilibrium?

Two metal spheres each $1.0 \mathrm{cm}$ in radius are far apart. One sphere carries 38 nC, the other -10 nC. (a) What's the potential on each? (b) If the spheres are connected by a thin wire, what will be the potential on each once equilibrium is reached?
(c) How much charge moves between the spheres in order to achieve equilibrium?

Key Concepts

Charge Redistribution in Connected Conductors

When two or more conductors are connected by a thin wire or any conductive path, they reach electrostatic equilibrium by redistributing their charges until they attain the same electric potential. This process occurs because charges move in response to differences in potential, and establishing equal potential minimizes the system’s overall energy, forming a foundation for understanding circuits and electrostatic behavior.

Conservation of Charge

The principle of conservation of charge states that the total electric charge in an isolated system remains constant regardless of how the charge is distributed among its parts. This concept is crucial when analyzing processes like connecting conductors, as it ensures that the sum of the charges before redistribution equals the sum after redistribution, providing a constraint to calculate how charge is shared between the conductors.

Electric Potential of a Charged Conductor

The electric potential on the surface of a charged conductor is determined by the charge on the conductor and its geometry. In the case of a sphere, the potential can be derived using the formula V = kQ/R, where k is Coulomb’s constant, Q is the total charge, and R is the radius of the sphere. This concept is fundamental in electrostatics as it shows how charge and geometry influence the potential at a point in space.

Capacitance of a Sphere

Capacitance is a measure of a conductor’s ability to store charge per unit potential, and for an isolated sphere, it is given by the formula C = 4???R, where ?? is the vacuum permittivity and R is the radius of the sphere. This intrinsic property of the sphere plays a key role in determining how much charge is needed to reach a given potential, thereby linking it to the sphere’s geometric characteristics.

Transcript

00:01 Let's start with the explanation of this particular problem.

00:04 We'll try to jot down the data first.

00:08 Here, there are two spears.

00:11 So let's just draw them.

00:15 The radius of four spear is given as r1 and the radius of second spear is taken as r2.

00:24 The charge on the first one is given and the charge on the second one is q2.

00:30 Now let's jot down the data that is r1 is equal to r2 which is given as 1 centimeter to convert into a meter it is going to be 0 .01 meter the value of q1 i'll mention here the charge on the first spear is given as 38 into 10 x to minus 9 columns that is 38 nanolombs and the value of q2 that is the charge on the second sphere is given as minus 10 into 10 to minus 9 columns.

01:18 Now let's start with the solution here as per given three different questions.

01:24 So for step a, we need to find potential on each sphere.

01:33 To start with, let the potential of the first spear be v1.

01:48 So therefore, v1 is equal to k times q1 upon r1.

01:56 So the value of k is 1 upon 4 pi epsilon not, q1 will remain same, multiplied by r1.

02:08 So therefore, v1 is equal to, let's put.

02:11 Put the value 0, that is 38 into 10 to minus 9 columns, divided by 4 pi, put the value of epsilon not as 8 .854 into 10 to minus 12 ferrads per meter, and the value of r1 as 0 .01 meter.

02:33 Let's go to the next page, therefore, value of v1 is coming as 34 ,153 volts.

02:46 So therefore v1 is equal to 34 .15 kilowatts.

02:55 And this is going to be one of the answers.

03:00 So i'll just mention here.

03:05 Okay, the next one.

03:07 Let the potential of the second sphere be v2.

03:17 And let's calculate that using the same formula.

03:20 That is k times q2 upon r2 put the value of k as 1 upon 4 pi absalom not which is the permittivity multiplied by r2 and now we'll put the values here so therefore v2 is equal to minus 10 into 10 less to minus 9 columns multiplied by 8 .854 into 10 to minus 12 ferrads per meter and 0 .01 meter as well.

04:01 So therefore it is going to be minus of 8 ,987 volts.

04:08 Therefore we can write the we do as minus of 8 .98 kilowatts...

Please give Ace some feedback