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Physics Lesson 14.6.3 - Electric Flux of a Charged Sphere. Gauss Law
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Welcome to our Physics lesson on Electric Flux of a Charged Sphere. Gauss Law, this is the third lesson of our suite of physics lessons covering the topic of Electric Flux. Gauss Law, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Electric Flux of a Charged Sphere. Gauss Law
Let's compute the electric flux across a spherical surface of radius R that contains a nearly point charge Q at its centre.
As we have explained earlier, electric field is the same in all points on the surface of the sphere. It is produced by the point charge Q and the field lines spread outwards (when Q is positive) from the charge to the surface of the sphere, in the perpendicular direction to the surface at any point. Therefore, the perpendicular component of the electric field summed across the entire surface is simply the electric field at a distance R from the charge multiplied by the area of the surface. In this case, the field vector is in the direction of area vector at every point. This means the angle between them is zero and cos 0° = 1. Thus,
= E × A × cos 00
= E × A
= 1/4πϵ0 × Q/R2 × 4πR2
Thus, we obtain for the electric flux Φ of a charged sphere:
Here we have made use of the equation of the area of a sphere A = 4πR2 and that of electric field of a point charge E = Q/4πϵ0R2.
The last formula is one of the many representations of the Gauss Law, which in simple word says:
The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface.
The sphere we considered above is called Gaussian Sphere. As you see, the electric flux does not depend on the radius of sphere but only on the amount of charge it carries at its centre.
Example 3
A -7μC point charge is placed at centre of a sphere whose radius is equal to 50 cm. Calculate the electric flux on the surface of the sphere.
Solution 3
The flux will be negative as the charge is negative. This means the electric field lines will be in the inward direction.
From the Gauss Law, we see that electric flux of a sphere is independent from the radius R. Thus, we have
= -7 × 10-6 C/8.85 × 10-12 C2/N × m2
= -7.9 × 105 V × m
You have reached the end of Physics lesson 14.6.3 Electric Flux of a Charged Sphere. Gauss Law. There are 4 lessons in this physics tutorial covering Electric Flux. Gauss Law, you can access all the lessons from this tutorial below.
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