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.
Answer:
(a) -472.305 J
(b) 1 m
Explanation:
(a)
Change in mechanical energy equals change in kinetic energy
Kinetic energy is given by
Initial kinetic energy is
Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero
Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence
0-472.305 J=-472.305 J
(b)
From fundamental kinematic equation
Where v and u are final and initial velocities respectively, a is acceleration, s is distance
Making s the subject we obtain
but a=\mu g hence
Answer:
c
Explanation:
a vector quantity has both magnitude and direction