Two charges, +9 µC and +16 µC, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the ne
t force (in N) on a −7 nC charge when placed at the following locations.
(a) halfway between the two
(b) half a meter to the left of the +9 µC charge
(c) half a meter above the +16 µC charge in a direction perpendicular to the line joining the two fixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the +x-axis.)
(a) halfway between the two
(b) half a meter to the left of the +9 µC charge
(c) half a meter above the +16 µC charge in a direction perpendicular to the line joining the two fixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the +x-axis.)
1 answer:
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To solve this problem we will apply the concepts related to the kinematic equations of motion. We will start calculating the maximum height with the given speed, and once the total height of fall is obtained, we will proceed to calculate with the same formula and the new height, the speed of fall.
The expression to find the change in velocity and the height is,
Replacing,
Thus the total height reached by the ball is
H = 22m+13.0612m
H = 35.0612m
Now calculate the velocity while dropping down from the maximum height as follows
Substituting the new height,
First we have to calculate the time taken to travel the distance 30 m, is
.
Now from equation of motion,
Given, .
As object starts from rest, so u = 0.
Substituting these values in above equation, we get
.
Thus, the acceleration is