Question

Problem Statement A 5.0-u point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-u point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero? Visual Representation • Draw a sketch of the charge distribution. • Establish a coordinate system and show the locations of the charges. Identify the point P at which you want to calculate the electric field. • Draw the electric field of each charge at point P. • Use symmetry to determine if any components of Ener are zero.

Answer 1

Given the charge q1 = 5$\mu$C loacted at 0cm of a meter stick and q2 = -4$\mu$C loacted at 50cm of the meter stick. So the distance between the two charges is r = 50-0 = 50cm. The diagram showing the arranagement is given below.

น q2 P Oon Socm * A Y -

Since the charges are of opposite sign, the point where the net field becomes zero will not lie in-between the charges but will lie ouside the charges, nearer to the charge having smaller magnitude. So here, the point where the net field is zero lies near q2. Let that point be P and that distance be x from q2.

The electric field at P due to q2 is,

$E_{2}=k\frac{q_{2}}{x^{2}}$

The electric field at P due to q1 is,

$E_{1}=k\frac{q_{1}}{(r+x)^{2}}$

The net field at P is zero. So,

$E_{1}=E_{2}$

$k\frac{q_{1}}{(r+x)^{2}}=k\frac{q_{2}}{x^{2}}$

$\frac{q_{1}}{(r+x)^{2}}=\frac{q_{2}}{x^{2}}$

$\frac{5\times 10^{-6}}{(0.5+x)^{2}}=\frac{4\times 10^{-6}}{x^{2}}$

$\frac{2.24}{0.5+x}=\frac{2}{x}$

$2.24x=2(0.5+x)$

$2.24x=1+2x$

$2.24x-2x=1$

$0.24x=1$

$x=\frac{1}{0.24}=4.17m=417cm$

So the electric field is zero at point 417cm from q2. Therefore the point is 417+50 = 467cm from 0cm.