The dome of a Van de Graaff generator receives a charge of 2.0 ×10^-4 C. Find the strength of th

The dome of a Van de Graaff generator receives a charge of...

The dome of a Van de Graaff generator receives a charge of $2.0 \times 10^{-4}$ C. Find the strength of the electric field (a) inside the dome, (b) at the surface of the dome, assuming it has a radius of $1.0 \mathrm{m},$ and $(\mathrm{c}) 4.0 \mathrm{m}$ from the center of the dome. Hint: Sce Section 15.5 to revicw propertics of conductors in electrostatic equilibrium. Also, note that the points on the sur- face are outside a spherically symmetric charge distribution; the total charge may be considered to be located at the center of the sphere.

The dome of a Van de Graaff generator receives a charge of $2.0 \times 10^{-4}$ C. Find the strength of the electric field (a) inside the dome, (b) at the surface of the dome, assuming it has a radius of $1.0 \mathrm{m},$ and $(\mathrm{c}) 4.0 \mathrm{m}$ from the center of the dome. Hint: Sce Section 15.5 to revicw propertics of conductors in electrostatic equilibrium. Also, note that the points on the sur-
face are outside a spherically symmetric charge distribution; the total charge may be considered to be located at the center of the sphere.

Key Concepts

Gauss's Law

Gauss's Law states that the net electric flux through a closed surface is proportional to the charge enclosed by that surface. This law is key in calculating electric fields for symmetrical charge distributions by selecting an appropriate Gaussian surface, simplifying the computation of the field based on the symmetry of the problem.

Electrostatic Equilibrium in Conductors

In electrostatics, conductors reach a state of equilibrium where the electric field inside the material is zero. Charges reside on the surface of a conductor, and this fact significantly influences the electric field distribution both inside and on the surface of the conductor.

Spherical Symmetry in Charge Distributions

A spherically symmetric charge distribution means that the charge is uniformly distributed over the surface of a sphere or within a spherical volume. This symmetry allows one to treat the sphere as if all its charge were concentrated at its center, which simplifies the calculation of the electric field at points outside the sphere.

Electric Field of a Point Charge

The electric field produced by a point charge is given by Coulomb’s law, where the field magnitude decreases as the inverse square of the distance from the charge. This principle is applied for points outside a spherically symmetric charge distribution by treating the distribution as if all the charge were located at the sphere’s center.

Transcript

00:01 Hello students let's start our discussion a vendor graph generator when the graph generator suppose this is the dome of our vendor graph generator okay, now we have to find out the electric field at various places of this wendegrave generator firstly we have to find out the electric field on a point that lies inside this wendegraff generator inside this dome.

00:34 Okay, so basically electric field at point p that is inside will basically the potential at point p that lies inside will be 1 upon 4 pi epsilon node q by r where r is the radius of this dome and q is the total charge that lies on the surface of the dome.

01:04 This is the cube.

01:07 Now, you are wondering that why i am saying the potential inside by taking the disk charge.

01:17 Basically, by the vendegraff generator properties and principle, the potential lies at the surface is exactly equal to potential lies at the inside.

01:29 Okay, so vp inside is 1 upon 4 pi epsilon node q by r so electric field p will also v by r that is so it will be 1 upon 4 pi epsilon node q by r square okay, so by solving this if the value of charge q is 2 into 10x bar minus 4 coulum and the radius of the dome is 1 meter so by putting the values 9 into 10 riss power 9 into 2 into 10 this by minus 4 by 1 square we will get an answer of 1 .8 into 10 race power 6 newton per coulum see at point p lies inside so this electric field is exactly similar on the surface so at point p dash which lies on the surface of the dome, electric field is equal to 1 .8 into 10 raise par 6 newton per culum as because the potential at the surface and the inside are exactly equal.

02:50 So the electric field at surface and inside is exactly equal.

02:56 So this is that point p dash which lies on the surface.

03:01 Now suppose we have to find out electric field.

03:03 Field on another point that lies outside this dome...

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