A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of
1 answer:
You might be interested in
Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:
DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation
according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero
solving for x we get
x = 0.200 m
b)electric field at half way mean x =0.25
E = 3.84*10^{-4} N/C
<h2>Hello!</h2>
The answer is: 19.59 m
<h2>Why?</h2>Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.
When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:
We will use the following formula:
Where:
Vf= Final velocity = 0
Vi= Initial velocity =
g = Gravity Acceleration =
s = Traveled distance
Have a nice day!