Physics, 14.09.2019 09:30 malikbryant2002

Acharge located at (3,-5) in the x-y plane exerts an attractive force of 0.75n on another charge located at (2,7) (all distances are in cm) a) in what direction does this force point? the answer should be given as a counterclockwise angle with respect to the positive x-axis. b) write down an expression for this force in terms of unit vectors i and (don't forget your units). c) if the charge at (3,-5) is +2pic, determine the magnitude and sign of the other charge.

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Answer from: lilacastro

(a): $247.77^\circ.$

(b): $\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}.$

(c): Magnitude = $0.604\ C.$

Sign = negative.

Explanation:

(a):

Given that first charge is located at (3,-5) and second charge is located at (2,7).

The electrostatic force between two charges acts long the line joining the two charges, therefore, the direction of electrostatic force of attraction between these two charges is along the position vector of charge at (2,7) with respect to the position of charge at (3,-5).

Assuming, $\hat i,\ \hat j$ are the unit vectors along positive x and y axes respectively.

Position vector of charge at (3,-5) with respect to origin is

$\vec r_1 = 3\hat i+(-5)\hat j$

Position vector of charge at (2,7) with respect to origin is

$\vec r_2 = 2\hat i+7\hat j$

The position vector of at (2,7) with respect to the position of charge at (3,-5) is

$\vec r = \vec r_2-\vec r_1\\=(3\hat i-5\hat j)-(2\hat i+7\hat j)\\=1\hat i-12\hat j.$

If $\theta$ is the angle this position vector is making with the positive x axis then,

$\rm \tan\theta =\dfrac{y\ component\ of\ (\hat i-12\hat j)}{x\ component\ of\ (\hat i-12\hat j}=\dfrac{-12}{1} = -12.\\\theta =\tan^{-1}(-12)=-85.23^\circ.$

The negative sign indicates that this position vector is $85.23^\circ.$ below the positive x axis, therefore, in counterclockwise direction from positive x axis, its direction = $360^\circ-85.23^\circ = 247.77^\circ.$

The force is also along the same direction.

(b):

According to Coulomb's law, the expression of this force is given by

$\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \hat r$

where,

$q_1,\ q_2$ are the charges. $\vec r$ = position vector of one charge with respect to another charge. $\hat r$ = unit vector along the direction of $\vec r$ .

= Coulomb's constant, which have value = $\rm 9\times 10^9\ Nm^2/C^2$ .

We have,

$\vec r = \hat i-12\hat j\\\therefore |\vec r| = \sqrt{1^2+(-12)^2}=\sqrt{145}.\\\Rightarrow \hat r = \dfrac{\vec r}{|\vec r|} = \dfrac{\hat i-12\hat j}{\sqrt{145}}$

Putting this value, we get,

$\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}$

(c):

If charge at (3,-5) is $+2\ pC$ , then the magnitude of the force between the two charges is given by

$F=k\ \dfrac{q_1q_2}{|\vec r|^2}$

where, we have,

$F=0.75\ N\\|\vec r| = \sqrt{145}\ cm = \sqrt{145}\times 10^{-2}\ m.\\q_1 = +2\ pC = +2\times 10^{-12}\ C.$

Putting all these values,

$0.75=9\times 10^9\times \dfrac{(+2\times 10^{-12})\times q_2}{(\sqrt{145}\times 10^{-2})^2}\\\Rightarrow q_2 = \dfrac{0.75\times(\sqrt{145}\times 10^{-2})^2}{9\times 10^9\times 2\times 10^{-12}} =0.604\ C.$

Since, the electric force between the two spheres is given to be attractive therefore this charge must be negative as the other charge is positive.

Answer from: Quest

Efficiency is an action of doing something. for example when you take a test. good efficiency means you did well but bad effeciency or no effiency means you did bad.

Answer from: Quest

the unit of the measure = $\frac{j}{kgk}$