The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.

at r < R:

Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.

E = 0.

at R < r < 2R:

The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.

at 2R < r:

The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.

Explanation:

Gauss’ Law is straightforward when applied to spheres. The area of the sphere is $A = 4\pi r^2$ , and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.

3 0

3 years ago

In the rectangle to the left, if

Margarita [4]

perimeter of a rectangle = 2(L+B)

90=2(L+B)

90/2=L+B

45=L+B

7 0

3 years ago

A solid, uniform disk of radius 0.250 m and mass 45.2 kg rolls down a ramp of length 5.40 m that makes an angle of 17.0° with th

serious [3.7K]

Explanation:

Given that,

Radius of the disk, r = 0.25 m

Mass, m = 45.2 kg

Length of the ramp, l = 5.4 m

Angle made by the ramp with horizontal, $\theta=17^{\circ}$

Solution,

As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :

(a) $mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$