Given:
10^10 electrons per second
To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:
From literature,
1 Coulomb is equivalent to 6.242×10^18 electrons<span>.
So,
= 10^10 electrons * (1 coulomb/</span><span>6.242×10^18</span> electrons) / second
<span>= 1.602 x 10^-9 coulumbs
This value is too small to be used in an actual setting.
</span><span>
</span>
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
brainly.com/question/2004529
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Answer:
The magnitude of the flux of electric field through a square of surface area is zero.
Explanation:

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

Well, if your question is how light affects plants,
then you would want to design an experiment that plays aruond with the amount of light a plant gets
thus the thing changing (or variable) would be amount of light