A high diver dives off a platform and hits the water 3.6 seconds later. How high was the platform?
2 answers:
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Answer:
The electric potential at point B is 83.33 V
The magnitude of the electric field at point B is 166.67 N/C
Explanation:
Given;
charge of an object, q = −6.00×10⁻⁹ C
Kinetic energy = 5.00×10⁻⁷ J
Electric potential at point A = +30.0 V
To determine the electric potential at point B, we apply the following formula;
Fd = qEd
Also, Fd is work done in moving the charge to 0.5 m = K.E
Again, Ed is the electric potential at point B (VB)
Substitute for V and K.E in the above equation, we will have;
K.E = q(VB)
VB = K.E/q
To determine the magnitude of the electric field at B, we use equation below;
V = Ed