Answer:
E = k*Q₁/R₁² V/m
V = k*Q₁/R₁ Volt
Explanation:
Given:
- Charge distributed on the sphere is Q₁
- The radius of sphere is R₁
- The electric potential at infinity is 0
Find:
What is the electric field at the surface of the sphere?E.
What is the electric potential at the surface of the sphere?V
Solution:
- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.
- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F = k*Q₁/R₁²
- Then the electric field at that point is
E = F/1
E = k*Q₁/R₁² V/m
- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V = k*Q₁/R₁ Volt
Well for question 1, the answer would be your opinion on which graph is easier to read and understand better.
for the second question. the one to get specific data would be the table chart because you can read the numbers easily rather than trying to guess on the bar chart wear the bars are actually at are kinda close to an exact number, but not really the number you were thinking
Answer:
What is 2^-3 x 3^-2 as a fraction?
What is (-2)^-3 x (-3)^-2 as a fraction
Explanation:
I need these pls. SORRYYY
-- loud sounds
-- bright lights
-- strong radio signals
-- Slinkies that can pinch you painfully
-- a tsunami in the ocean
-- earthquakes above Richter 5 or 6
