Answer:
Check the attached image below.
Explanation:
Kindly check the attached image below to get the step by step explanation to the above question.
Ummm because it need space to move around and function
Not really the volume of a container is simply length X width X depth so just how big the container unless the water is pressurized by some sort of weight or if the containers air pressure is lowered
Answer:
Ex(P) = -3.602 x 10^6 N/C
Explanation:
- q2 = 2.6 μC = the net charge on the conducting shell
- inner radius of conducting shell = a = 2.2 cm =0.022m
- outer radius of conducting shell = b = 4.5 cm = 0.045m
1) To get Ex(P), the value of the x-component of the electric field at point P, located a distance 8.8 cm along the x-axis from q1 ;
Ex(P) = k(q1+q2)/r^2
= 9 x 10^9 (-5.7 + 2.6) x 10^-6 /0.088^2
Ex(P) = -27.9 x 10^3/ 0.007744
Ex(P) = -3.602 x 10^6 N/C
