Question

QUESTION 4 a2 21 Not enough information to

Answer 1

To study this problem we will draw a free-body diagram for the left charge. The charge is in equilibrium under the application of the forces $\vec{T}$ of the wire, the electric force$\vec{F}_{21}$ coming from the other load and the gravitational force $m_{1}\vec{g}$.

Because the load is in equilibrium, each of the forces in the horizontal and vertical directions must add zero.

(1) $\sum F_{x}=Tsin\theta -F_{21}=0$

(2) $\sum F_{y}=Tcos\theta -m_{1}g=0$

From equation (2), we see that $T=m_{1}g/cos\theta$ ; then $T$can be substituted for equation (1) using that value. With this we obtain a value for the magnitude of force $F_{21}$:

$F_{21}=m_{1}gtan\theta =(0.23g)(9.8m/s^{2})tan(15.9^{0})$

$F_{21} =(0.00023kg)(9.8m/s^{2})tan(15.9^{0})$

$F_{21} =6.42x10^{-4}N=6.42E-4N$

According to Coulomb's law, the magnitude of the electric force is equal to

$F_{21} =k_{e}\frac{q_{1}q_{2}}{r^{2}}$

Solving for $q_{1}$, we get

$q_{1}=\frac{F_{21}r^{2}}{k_{e}q_{2}}$

$q_{1}=\frac{(6.42x10^{-4}N)(2.5cm)^{2}}{(8.99N.m^{2}/C^{2})}$

$q_{1}=\frac{(6.42x10^{-4}N)(0.025m)^{2}}{(8.99N.m^{2}/C^{2})}$

$q_{1}=\frac{(6.42x10^{-4}N)(6.25x10^{-4}m^{2})}{(8.99N.m^{2}/C^{2})}$

$q_{1}=\frac{4.0125x10^{-7}N.m^{2}}{17.98N.m^{2}/C}$

$q_{1}=+2.23x10^{-8}C=+2.23E-8C$