Question

Four positive charges, each of 10.0 nC occupy four corners of a square with sides of 0.10 m. a) Find the magnitude of the total electrostatic force on charge Q1 due to other three charges? b) Find the orientation (the angle with respect to positive x axis) of the total electrostatic force on charge Q1 due to other three charges? Ql. Q2 0.1m Q3 Q4

Answer 1

Given that each charge is +10nC and the side of the square is a = 0.1m. The diagram of the forces acting on the charge Q1 is shown below in the figure.

Fik 13 a Fiz Q₂ N. A > 10 03

(a) The magnitude of force acting on Q1 due to Q2 is,

$F_{12}=k\frac{Q_{1}Q_{2}}{a^{2}}$

$k=9\times 10^{9}Nm^{2}/C^{2}$

$F_{12}=9\times 10^{9}\times \frac{10\times 10^{-9}\times 10\times 10^{-9}}{(0.1)^{2}}=9\times 10^{-5}N$

Since Q1 and Q2 are positive. the direction of this force is away from Q2 as shown in the figure.

The magnitude of force acting on Q1 due to Q3 is,

$F_{13}=k\frac{Q_{1}Q_{3}}{a^{2}}$

$F_{13}=9\times 10^{9}\times \frac{10\times 10^{-9}\times 10\times 10^{-9}}{(0.1)^{2}}=9\times 10^{-5}N$

Since Q1 and Q3 are positive. the direction of this force is away from Q3 as shown in the figure.

The magnitude of force acting on Q1 due to Q4 is,

$F_{14}=k\frac{Q_{1}Q_{4}}{r^{2}}$

$r^{2}=a^{2}+a^{2}=(0.1)^{2}+(0.1)^{2}=0.02$

$F_{14}=9\times 10^{9}\times \frac{10\times 10^{-9}\times 10\times 10^{-9}}{0.02}=4.5\times 10^{-5}N$

Since Q1 and Q4 are positive. the direction of this force is away from Q4 as shown in the figure.

The resultant of F12 ans F13 is given by,

$F=\sqrt{(F_{12})^{2}+(F_{13})^{2}}$

$F=\sqrt{(9\times 10^{-5})^{2}+(9\times 10^{-5})^{2}}=12.73\times 10^{-5}N$

The direction of this resultant clockwise from negative x-axis is

$\theta' = \tan^{1}(\frac{F_{13}}{F_{12}})=\tan^{-1}(\frac{9\times 10^{-5}}{9\times 10^{-5}})=\tan^{-1}(1)=45\degree$

Since its a square, the angle $\theta$ = $45\degree$ . That means F and F14 are in the same direction. Therefore, the magnitude of the total force on Q1 by the other three charges is,

$F_{T}=F+F_{14}=12.73\times 10^{-5}+4.5\times 10^{-5}=17.23\times 10^{-5}N$

So the magnitude of the total electrostatic force on Q1 due to other charges is 17.23 x 10^-5N.

(b) The orientation of this total electrostatic force clockwise from negative x-axis is $45\degree$ . Therefore the orientation of this total electrostatic force anti-clockwise from positive x-axis is,

$\alpha = 180\degree-45\degree=135\degree$

So the orientation of the total electrostatic force on Q1 due to other charges with respect to positive x-axis is $135\degree$ .