I'm unsure what to do about this Coulomb's Law problem. Help?
a) Find the net Coulomb force in vector form on charge q1, due to the presence of charges on q2 and q3.
I did my best to type the "square" but the "o" on the top left is q1 (-3.0*10^-6) and below it is q3 (3*10^-6). The "o" at the bottom right is -3.0*10^-6.
b) If charges q2 and q3 were nailed in place, find the strength of the acceleration of charge q1, which has a mass of 12 g (0.012 kg).
Also the original problem: https://i.postimg.cc/4NFfKvmY/20190622-171246.jpg
I know that the vectors need to be divided up to find what the hypotenuse would be. At first, I just plugged the values into Coulomb's formula: Fc = k |q1*q2| / r^2
- Steve4PhysicsLv 76 months agoFavorite Answer
The force on q1 due to q3 (r=10cm = 0.1m) is:
F13 = k.q1.q3/r² = 9*10^9 * (-3*10^-6) * (3*10^-6) / 0.1² = -8.1N
The minus sign means attraction. So q1 is being pulled towards q3 which is the minus y direction. There is no x component. This means F13 has:
x component = 0
y component = -8.1N
The diagonal (d) of the square is given by Pythagoras: d² = 0.1² + 0.1² = 0.02
The force on q1 due to q2 (with r=d so r² = d² = 0.02) is:
F12 = k.q1.q1/r² = 9*10^9 * (-3*10^-6)*(-3*10^-6)/0.02 = 4.05N
This is positive meaning repulsion. So q1 is being push away (up/left) from q2 along the square’s diagonal: so the force has a positive y component and a negative x component.
x component = -4.05cos(45º) = -2.8N
y component = 4.05sin(45º) = 2.8N
The resultant force on q1 is the vector sum F = F13 + F11:
total of x components = 0 + (-2.8) = -2.8N
total of y components = 8.1 + 2.8 = 10.9N
So in vector format: F = -2.8Nî + 10.9Nĵ (sometimes written <2.8N, 10.9N>).
Check my arithmetic of course.
Find the magnitude of F using |F|= √((-2.8)² + 10.9²).
(Though it's best to use unrounded values to minimise rounding errors.)
Then acceleration a = |F|/m (remembering to convert m from grams to kg).
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