Question

When we find the electric field due to a continuous charge distribution, we imagine slicing that source up into small pieces, finding the electric field produced by the pieces, and then integrating to find the electric field. Let's see what happens if we break a finite rod up into a small number of finite partides. The figure below shows a rod of length 2 carrying a uniform charge Q modeled as five particles of charge Q/5. Two particles are at the ends of the rod, and the rest are evenly spaced along the rod. Find an expression for the electric field at point A located a distance & above the midpoint of the rod using each of two methods (to make your work easier, you can calculate any square root to three significant figures). Q/5 0/5 0/5 0/5 Q/5 (a) modeling the rod with just five particles (Use the following as necessary: &, Q. and k.) total.(a) - 0.627 kg X (b) using the exact expression (Use the following as necessary: 4, Q, and k.) Etotal.b) - 0.7070 22 -> (c) Compare your approximate result to your exact expression for the rod by finding the ratio of the approximate expression to the exact expression. (Enter the ratio of the magnitudes.) Etotal.al - D 8876 Exptal.(b)

Answer 1

VZE E E P ve А. B q t

As shown in figure if two charges each having same magntiude q are symmetrically placed on a horizontal line at A and B with respect to P so that distance l between point p to the horizontal line is same as the distance l between each charge to the centre of charges on the line AB.

Resultant field in the direction perpendicular to line AB is given by

ER = V2K-9 2 x 12

where K = 1/($4\pi \epsilon _{o}$) is coulomb constant

E P V2e V312 . V2e V3012 B A005 9 12 12 12 12) q 5 -05 5 5

when five equal charges q/5 is symmetrically placed on the line AB then electric field E1 is given by

..............................(1)

E = K Q. 5 x 12 1+ 2y2 + v2 = 0.459K2 3 4 12

If charge is distributed uniformly , exact electric field E is given by

$E = K \times \frac{Q}{\sqrt{2}l^{2}} = 0.707 K \times \frac{Q}{l^{2}}$......................................(2)

Hene ratio of approximate electric field to exact electric field is given by

$\frac{E_{1}}{E} = \frac{0.459}{0.707} \approx 0.65$

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Second question

Electric field E due to charge q

$E_{1} = K \frac{q}{y^{2}+\frac{d^{2}}{4}}$

Above field has direction along the line connecting charge q and point A. Hence this field can be written in vector form as

$E_{1} = \left ( K \frac{q}{y^{2}+\frac{d^{2}}{4}}sin\theta \right ) \hat{i} + \left ( K \frac{q}{y^{2}+\frac{d^{2}}{4}} cos\theta \right ) \hat{j}$

where $\theta$ is the angle made by field vector with y-axis. Hence above expression can be written as

$E_{1} = \left ( K \frac{q\times \frac{d}{2}}{(y^{2}+\frac{d^{2}}{4})^{3/2}} \right ) \hat{i} + \left (K \frac{q\times y}{(y^{2}+\frac{d^{2}}{4})^{3/2}} \right ) \hat{j}$

Similarly we can write electric field E2 due to charge q as

$E_{2} = \left ( K \frac{2q\times \frac{d}{2}}{(y^{2}+\frac{d^{2}}{4})^{3/2}} \right ) \hat{i} + \left (K \frac{2q\times y}{(y^{2}+\frac{d^{2}}{4})^{3/2}} \right ) \hat{j}$

Hence total electric field is given by

$E_{1}+E_{2} = \left ( K \frac{\frac{3}{2}q\times d}{(y^{2}+\frac{d^{2}}{4})^{3/2}} \right ) \hat{i} + \left (K \frac{3q\times y}{(y^{2}+\frac{d^{2}}{4})^{3/2}} \right ) \hat{j}$