Question

Problem 2 -Superposition Principle for electrie forces between point charges Two charges are positioned along the horizontal -axis. 6.0 uC is positioned in the origi 02 =-60 pC is positioned at x = 2.0 m. A third charge 03-3.0 pC is placed halfway betwe 01 and 02 (at the point x = 1.0 m). Compute: a) The magnitude of the net electric force acting on Q b) The direction of the net electric force acting on Q3. e) The electric field produced by the charges 01 and O2 at the position of charge Qs.

Answer 1

Solution:

Charge distibution is given below.

(o, 0) (2,0) 13 23

(a) Solution:

Force acting on charge Q3 due to Q2 is given by

$F_{23}=\frac{1}{4\pi \varepsilon _{o}}\frac{Q_{2}Q_{3}}{r^{2}}$

$F_{23}=9\times 10^{9}\times \frac{-6\times 10^{-6}\times 3\times 10^{-6}}{1^{2}}$

$F_{23}=-0.162\; \mathrm{N}$

Negative sign indicates that, the force toward chage 'Q₂'

Force on charge 'Q3' due to charge 'Q1'

$F_{13}=\frac{1}{4\pi \varepsilon _{o}}\frac{Q_{1}Q_{3}}{r^{2}}$

$F_{13}=9\times 10^{9}\times \frac{6\times 10^{-6}\times 3\times 10^{-6}}{1^{2}}$

$F_{13}=0.162\; \mathrm{N}$

this force is acting toward 'Q3' as indicated in above figure.

So, net force acting on 'Q3' is given by

$F_{net}=F_{13}+F_{23}$

$F_{net}=0.162+0.162$

$F_{net}=0.324\; \mathrm{N}$

(b) Solution:

The direction of the resultant force is towards position x-axis as shown in figure.

(c) Solution:

The electric field can be calculated as follows

$F_{net}=EQ_{3}$

$E=\frac{F_{net}}{Q_{3}}$

$E=\frac{0.324}{3\times 10^{-6}}$

$E=0.108\times 10^{6}\; \mathrm{v/m}$