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forsale [732]
1 year ago
5

Describe the equipotential surfaces for (b) a uniformly charged sphere.

Physics
1 answer:
storchak [24]1 year ago
8 0

Equipotential surfaces are nesting concentric spheres around a uniformly charged sphere.

<h3>What is the shape of equipotential surface for a uniformly charged sphere.</h3>

In mathematics and physics, an area of space is said to be equipotential or is potential if every point within it has the same potential. Although it also applies to vector potentials, this often refers to a scalar potential. Equipotential surface refers to the region where all points have the same potential. On the equipotential surface, a charge can be moved from one point to another without exerting any effort. In other terms, an equipotential surface is any surface that has the same electric potential at every point.

  • The surface where all the points have the same electric potential is known as the equipotential potential.
  • Equipotential surfaces have the shape of a sphere with a homogeneous charge in the centre surrounded by concentric spheres.

To learn more about equipotential surfaces, refer to:

brainly.com/question/14675095

#SPJ4

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3 years ago
Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
Naddik [55]

Answer:

4.30 x 10⁵ N/C

Explanation:

Two positive +3.0-μC point charges at opposite corners of the square creates equal and opposite electric field at the center. hence the electric field by these two positive charges at opposite corners becomes zero.

a = length of the square of side = 0.50 m

r = distance of the center from each corner = \frac{a}{sqrt(2)} = \frac{0.50}{sqrt(2)} = 0.354 m

Magnitude of net electric field at the center is given as

E = \frac{2kq}{r^{2}} \\E = \frac{2(8.99\times10^{9})(3\times10^{-6})}{(0.354)^{2}} \\E = 4.30\times10^{5} NC^{-1}

6 0
3 years ago
A 1.7 kg book is lying on a 0.77 m -high table. You pick it up and place it on a bookshelf 2.0 m above the floor. how much work
luda_lava [24]
You are given the mass of the book with an initial height of 0.77m. You are also given the final height of the book, 2m above the floor. The problem requires you to find the work done on the book. It is said that work is equal to force times distance. And that force is equal to mass times acceleration due to gravity.
W=Fd
W=mgd=mgh
W=(1.7kg)(9.8m/s^2)(2m-0.77m)
W=20.49J
<span>The work is equal to 20.49J.</span>
4 0
3 years ago
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