Answer:
Where the electric potential is constant, the strength of the electric field is zero.
Explanation:
As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e
Eₓ = - dV / dx ----------(i)
From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.
<em>Therefore, a constant electric potential means that electric field is zero.</em>
Answer: 14. 49 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
Where:
is the horizontal distance between the cannon and the ball
is the cannonball initial velocity
since the cannonball was shoot horizontally
is the time
is the final height of the cannonball
is the initial height of the cannonball
is the acceleration due gravity
Isolating from (2):
(3)
(4)
(5)
Substituting (5) in (1):
(6)
Finally:
