Answer to Question #158905 in Electricity and Magnetism for Anand

Question

Two positively charged particles each having charge 20 nC, are kept at a distance

of 2.0 m from each other. Determine the electric field due to each charge. Show

the electric field vectors on an appropriate diagram.

Answer 1

Explanations & Calculations

• The electric field intensity is defined as the force imposed on a unit positive charge due to the other surrounding electric charges.
• It is a vector quantity then & during a drawing of field intensity, that should be considered.
• By Coulomb's law, the magnitude of the intensity due to an isolated charge is given by

$\qquad\qquad \begin{aligned} \small |\bold{\overrightarrow E}|&= \small \frac{1}{4\pi \epsilon}\frac{|Q|}{r^2} \end{aligned}$ direction is in line with the positive unit charge

• Therefore, from a single of the given charges, the generated field magnitude at an arbitrary point is

$\qquad\qquad \begin{aligned} \small |\bold{\overrightarrow E_1}|&= \small \frac{1}{4\pi \epsilon}\frac{(+20nC)}{r_1} \end{aligned}$

• When both the charges are considered,

$\qquad\qquad \begin{aligned} \small |\overrightarrow E_{net}|&= \small |\overrightarrow E_1|+|\overrightarrow E_2|\\ &= \small \frac{1}{4\pi \epsilon}\frac{20}{r_1}+\frac{1}{4\pi \epsilon}\frac{20}{r_2}\\ &= \small \frac{20}{4\pi \epsilon}\Big[\frac{1}{r_1}-\frac{1}{r_2}\Big] \end{aligned}$

• And it is the vector addition which gives the direction of the field at that point.

• The distribution of the field vectors of the two charges can be sketched as follows

• Each arrowed line shows the behaviour of the field intensity creating a zero intensity region right in the middle (1m from each) of the two as they are identical charges.
• Note how the vectors are added (roughly) at the bottom right.

• The distribution of the magnitude of the intensity is as follows