Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po
1 answer:
Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
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The downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.
<h3>Downward acceleration of the cylinder</h3>
The downward acceleration of the solid cylinder is determined from the principle of conservation of angular momentum as shown below;
Iα = Tr
where;
- I is moment of inertia of the solid cylinder
- α is angular acceleration of the cylinder
- T is tension in the string
- r is length of the string
α = Tr/I
where;
- a is the downward acceleration of the solid cylinder
- R is radius of the cylinder
Thus, the downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.
Learn more about acceleration here: brainly.com/question/605631
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