Answer:
.
Explanation:
Since no external force is acting on the system.
Therefore, Total energy remains constant before and after.
So, Total energy of system= energy due to potential applied+kinetic energy

(Here v=velocity ,V=potential ,q=charge and m=mass).
Putting values .
We get,
.
At point B charged particle is moving faster as compared to point A.
Hence, it is the required solution.
Answer:
a = 0.5 m/s²
Explanation:
Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:
ωf = ω₀ + α*t
⇒ ωf = α*t ⇒ α =
= 
The angular velocity, and the linear speed, are related by the following expression:
v = ω*r
Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:
a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²
Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>