# Electricity and Gravity

1. Four charges, q1 = +200 µC, q2 = +54 µC, q3 = -134 µC, and q4 = +33 µC, are fixed at the corners of a 4 m by 5 m rectangle, as illustrated in the figure below. What are the magnitude and the direction of the net force acting on q1? (Assume the x-axis extends from q1 to the right.). Distance between q1 and q4 is 5 m, distance between q4 and q3 is 4 m, distance between q3 and q2 is 5 m and distance between q2 and q1 is 4 m.

2. At what point between Earth and Mercury is the net force of gravity on a body by both Earth and Mercury exactly zero?
$ME = 6.00 *10^{24} kg$,
mass of Mercury
$MM = 3.30 *10^{23} kg$,
and the distance between their centres
$REM = 7.73 *10^7 m?$
1. X axis from q1 to the right, y axis from q1 downwards
$F(4,1) = q1*q4/(4 \pi \epsilon_0)/R14$
$1/(4 \pi \epsilon0) =9*10^9$
$F41 = +33*200*10^{-12}*9*10^9/5 =11.88 N$ along the positive direction of x (to the right)
$F21 =q2*q1/(4 \pi \epsilon0)/R21 =54*200*10^{-12}*9*10^9/4 = 24.3 N$ along with the positive direction of Y (downwards)
$R31 =\sqrt{(R21^2+R41^2)} =\sqrt{(16+25)} =6.4 m$
$F31 = q1*q3/(4 \pi \epsilon_0)/R31 =-134*200*10^{-12}*9*10^9/6.4 =-37.69 N$ from q1 to q3.
Now one need to decompose F31 along x and y axis. alpha is the angle between F31 and x axis
$tan (\alpha) =4/5$, $\alpha =38.66 degree$
$F31x = F31*cos(\alpha) =-37.69*cos(38.66) =-29.43 N$ in the negative direction of x
$F31y = F31*sin(\alpha) = -37.69*sin(38.66) =-23.54 N$ in the negative direction of y
Now the total force is
$Fx = F(4,1) +F(3,1)x =11.88-29.43 = -17.55 N$
$Fy = F(2,1) +F(3,1)Y =24.3 -23.54 = +0.76 N$
Total force on charge 1 is
$F = \sqrt{(Fx^2+Fy^2)} =17.57 N$
let $\beta$ the angle of F with x axis
$tan(\beta) = Fy/Fx =-0.76/17.55 =-87.52$ degree with respect to the positive direction of x

2. the total gravitational force due to Mercury is equal to the gravitational force due to Earth
$MM/(REM-R12)^2 =ME/R12^2$
where R12 is the distance from Earth
$MM/ME = (REM-R12)^2/R12^2$
$0.055 = (REM/R12-1)^2$
$REM/R12 -1 =0.2345$
$REM/R12 =1.2345$
$R12 =REM/1.2345 =6.26*10^7 m$