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The equation is E=kq/r^2.

Would I use the net charge of the sphere, or do I need to calculate a new charge due to the positive one at the center?

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- Thread starter dpg1276
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- #1

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The equation is E=kq/r^2.

Would I use the net charge of the sphere, or do I need to calculate a new charge due to the positive one at the center?

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- #2

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For example, if the net negative charge on the conducting sphere is exactly equal to -1 times the positive charge in the middle, then the charges cancel, and there is no electric field outside the sphere.

To find out why it works this way, you could either do an integral and Coulomb's law, or use Gauss' law, using the symmetry of the problem.

The electric field of a point outside a charged sphere is the force experienced by a unit positive charge placed at that point. It is a vector quantity, meaning it has both magnitude and direction.

The electric field at a point outside a charged sphere can be calculated using the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance from the center of the sphere to the point.

The direction of the electric field at a point outside a charged sphere is radial, meaning it points directly away from or towards the center of the sphere. The direction depends on the charge of the sphere and the location of the point.

Yes, the electric field at a point outside a charged sphere is inversely proportional to the square of the distance from the center of the sphere. This means that the farther away the point is from the sphere, the weaker the electric field will be.

Yes, there are certain points outside a charged sphere where the electric field is zero. These points are called neutral points and they occur when the electric field vectors from different parts of the sphere cancel each other out.

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