Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
Hydroelectricity is the best answer.
This is an article by the EIA, but the pie graph is the most helpful: https://www.eia.gov/energyexplained/?page=us_energy_home
Answer:
Explanation:
Let the velocity of deuteron be v then force on it in magnetic field
Bqv , B is magnetic field and q is charge on deuteron . This force will provide centripetal force for circular path so
mv² / r = Bqv m is mass of deuteron and r is radius of circular path
v = Bqr / m
(.5 x 1.6 x 10⁻¹⁹ x 55.6 x 10⁻² )/ 3.34 x 10⁻²⁷
= 13.31 x 10⁶ m /s