Question

probability of getting more than a head in a die of two toss how to find electric field due to a point charge? derivition?

Answer 1

Answer:

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Answer 2

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.

Gauss Law is a powerful method of calculating electric fields. If you have a solid conducting sphere (e.g., a metal ball) that has a net charge Q on it, you know all the excess charge lies on the outside of the sphere. Gauss law tells us that the electric field inside the sphere is zero, and the electric field outside the sphere is the same as the field from a point charge with a net charge of Q. Thats a pretty neat result.

Fig. 1

The result for the sphere applies whether its solid or hollow. Lets look at the hollow sphere, and make it more interesting by adding a point charge at the center.

Fig. 2

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What does the electric field look like around this charge inside the hollow sphere? How is the negative charge distributed on the hollow sphere? To find the answers, keep these things in mind:

The electric field must be zero inside the solid part of the sphere

Outside the solid part of the sphere, you can find the net electric field by adding, as vectors, the electric field from the point charge alone and from the sphere alone

We know that the electric field from the point charge is given by kq / r2. Because the charge is positive, the field points away from the charge.

If we took the point charge out of the sphere, the field from the negative charge on the sphere would be zero inside the sphere, and given by kQ / r2 outside the sphere.

The net electric field with the point charge and the charged sphere, then, is the sum of the fields from the point charge alone and from the sphere alone (except inside the solid part of the sphere, where the field must be zero). This is shown in the picture:

Fig. 3