When vapor pressure equals atmospheric pressure, it's called boiling point of that liquid.
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>
Answer:
False
Explanation:
Please see the attached file
The value of the second charge is 1.2 nC.
<h3>
Electric potential</h3>
The work done in moving the charge from infinity to the given position is calculated as follows;
W = Eq₂
E = W/q₂
<h3>Magnitude of second charge</h3>
The magnitude of the second charge is determined by applying Coulomb's law.
Thus, the value of the second charge is 1.2 nC.
Learn more about electric potential here: brainly.com/question/14306881
