Question

1. Electric charge is distributed uniformly along a R thin rod of length a, with total charge Q. Take the у potential to be zero at infinity e a. Find the electric field Ē at point P, a distance x to the right of the rod (10 pts) b. Find the electric field Ē at point R, a distance y above of the rod (10 pts) c. In parts (a) and (b), what does your result reduce to as x or y becomes much larger than a?(5 pts) 2. Positive charge Q is distributed uniformly on the left quarter and negative charge Q is distributed on the right quarter of the semicircle of radius a, in Figure on the right (1) Calculate x and y component of the electric field at the center of the semicircle. (20 points) (ii) Find the Coulomb force acting on a negative charge placed at the center of the semicircle. (5 points)

Answer 1

1.a. First we consider that the point p is at a fixed distance d away from the edge of the rod and along the x axis. We can use the differential form of the E field definition

de dE=k- 2

If we consider the rod is made up of infinitesimal charges that contribute to the total field

9 d. ra k a E = 5 C + d)2

After integrating and evaluating the limits we get

$E=\frac{kQ}{a}\left ( \frac{1}{d}-\frac{1}{d+a} \right )$

Now we can assign the point p variable distance x along the horizontal axis band use vector notation to get the general expression:

$\bar{E}_{p}=\frac{kQ}{a}\left ( \frac{1}{x}-\frac{1}{x+a} \right )\hat{i}$

b. Similarly for point R if we first assume it's at a fixed distance d and take into consideration that now the E field has both x and y components we can write:

$E_{x}=\int_{0}^{a}k\frac{\frac{Q}{a}}{x^{2}+d^{2}}\frac{x}{\sqrt{x^{2}+d^{2}}}dx$

$E_{x}=\frac{kQ}{a}\int_{0}^{a}\frac{xdx}{\left (x^{2}+d^{2} \right )^{\frac{3}{2}}}$

After integrating and evaluating the limits we get:

$E_{x}=\frac{kQ}{a}\left ( \frac{1}{d}-\frac{1}{\sqrt{a^{2}+d^{2}}} \right )$

For the y component:

$E_{y}=\int_{0}^{a}k\frac{\frac{Q}{a}}{x^{2}+d^{2}}\frac{d}{\sqrt{x^{2}+d^{2}}}dx$

$E_{y}=\frac{kQd}{a}\int_{0}^{a}\frac{dx}{\left (x^{2}+d^{2} \right )^{\frac{3}{2}}}$

After integrating and evaluating the limits we get:

$E_{y}=\frac{kQ}{d}\left (\frac{1}{\sqrt{d^{^{2}}+a^{2}}} \right )$

Finally we can express the E field in vector form for a point R at a y distance from the edge as:

$\bar{E}_{R}=\frac{kQ}{a}\left ( \frac{1}{y}-\frac{1}{\sqrt{a^{2}+y^{2}}} \right )\hat{i}+\frac{kQ}{y}\left (\frac{1}{\sqrt{y^{^{2}}+a^{2}}} \right )\hat{j}$

c. When x becomes much larger than a we have:

$\bar{E}_{p}\approx 0$

$\bar{E}_{R}\approx \frac{kQ}{y^{2}}\hat{j}$

Please upload the second question separately. Thank you!