Question

A positive charge with a mass of 3 kg is placed on a string and a negative charge of the same amount (magnitude) but negative is placed directly to the right of this charge when in equilibrium. Also, the angle with the vertical of the string is 39.72 degrees when the charge is in equilibrium. Then the negative charge is moved directly below the charge of the string and its placed twice as far as it originally was to the charge on the string. What is the amount of tension in the string now in Newtons?

The string length is not given. Show all work please.

Answer 1

Applying Newton's second law on the positive charge after the negative charge is moved directly below the positive charge, we obtain

$\:$

$\sum F_{yAfter}=T_{2}-F_{2e}-mg=0$

$\Rightarrow \: \: \: T_{2}=F_{2e}+mg$

$\:$

where the magnitude of the electrical force F_2e is given by the eq.

$\:$

$F_{2e}=\frac{kq^{2}}{r_{2}^{2}}$

$\:$

thus,

$\:$

$T_{2}=\frac{kq^{2}}{r_{2}^{2}}+mg \: \: \: \: \: \: \: \: \: \: \: \: (1)$

$\:$

Now, we only have to find the magnitude of the charge q and the distance between the two charges r₂. We know that r₂ is twice as far as it originally was to the charge on the string. so

$\:$

$r_{2}=2r_{1}$

$\:$

Also, we can apply again Newton's second law to the positive charge, but this time before the negative charge was moved from its original point.

$\:$

$\sum F_{xBefore}=F_{1e}-T_{1x}=0$

$\:$

from geometry,

$\:$

$T_{1x}=T_{1}\sin \theta$

$\:$

thus,

$\:$

$F_{1e}-T_{1}\sin \theta =0$

$\Rightarrow \: \: \: \frac{kq^{2}}{r_{1}^{2}}-T_{1}\sin \theta =0 \: \: \: \: \: \: \: \: \: (2)$

$\:$

On the vertical axis, we have

$\:$

$\sum F_{yBefore}=T_{1y}-mg=0$

$\:$

from geometry,

$\:$

$T_{1y}=T_{1}\cos \theta$

$\:$

thus,

$\:$

$T_{1}\cos \theta -mg=0$

$\Rightarrow \: \: \: T_{1} =\frac{mg}{\cos \theta} \: \: \: \: \: \: \: \: \: \: \: \: (3)$

$\:$

inserting eq. (3) into (2), we obtain

$\:$

$\frac{kq^{2}}{r_{1}^{2}}-\left ( \frac{mg}{\cos \theta } \right )\sin \theta =0$

$\Rightarrow \: \: \: \: \frac{kq^{2}}{r_{1}^{2}}-mg \tan \theta =0$

$\Rightarrow \: \: \: \: \frac{kq^{2}}{r_{1}^{2}}=mg \tan \theta$

$\Rightarrow \: \: \: \: \frac{kq^{2}}{\left (\frac{r_{2}}{2} \right ) ^{2}}=mg \tan \theta$

$\Rightarrow \: \: \: \: \frac{4kq^{2}}{r_{2} ^{2}}=mg \tan \theta$

$\Rightarrow \: \: \: \: \frac{kq^{2}}{r_{2} ^{2}}=\frac{mg \tan \theta }{4} \: \: \: \: \: \: \: \: \: (4)$

$\:$

inserting eq. (4) into (1), we obtain

$\:$

$T_{2}=\frac{mg \tan \theta }{4}+mg$

$\Rightarrow \: \: \: T_{2}=mg\left (\frac{\tan \theta }{4}+1 \right )$

$\:$

inserting data given,

$\:$

$T_{2}=(3kg)(9.8m/s^{2})\left [\frac{\tan (39.72^{\circ}) }{4}+1 \right ]$

$\:$

or

$\:$

${\color{Blue} T_{2}=35.5\, N}$