A charge Q accumulates on the hollow metallic dome, of radius R, of a Van de Graaff generator. A probe measures the electric field strength at various points outside the sphere surface.
By what factor will the electric field value at the 2R distance be changed if the charge value were increased to (5/3)Q?

5/3

Explanation:

Assuming that the charge is evenly distributed on the surface of the sphere, by symmetry, at a same distance of the sphere, the electric field must be radial, and of the same magnitude.

So, if we define a spherical gaussian surface with radius equal to 2*R, we  can find the total flux as follows:

Flux = E*A = E*4*π*(2*R)² (1)

Now, Gauss' Law states that the flux we have just found must be equal to the charge enclosed, divided by the permitivitty of free space, ε₀:

Rearranging terms, and solving for E, we find the following expression:

As it can be seen, the electric field, at a distance 2R from the center, is just the same that it would produce a point charge with the same value of Qenc, located in the center of the sphere.

If we keep the same distance, and increase the charge value to 5/3 Q, we can easily see that the electric field (which is linear with the charge) will increase in the same factor, i.e. , 5/3.

⇒  

insects!

explanation:

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)