Two point charges have a total charge of 505 µC. When placed 1.10 m apart, the force each exerts on the other is 19.3 N and is repulsive. What is the charge on each? (Round your answers to the nearest µC.)
3 answers
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Let the two charges be Q1 and Q2.
In this case, Q2 = 505*10^-6 - Q1
Since the force is replusive, they both have the same sign.
Coulomb's Law tells you that
19.3 N = k*Q1*Q2/R^2, where R = 1.1 m
Substitute 505*10^-6 - Q1 for Q2 and solve for Q1.
You will have to look up k. It is something like 8.99*10^9 N(m^2/C^2), but I don't trust my memory.
In this case, Q2 = 505*10^-6 - Q1
Since the force is replusive, they both have the same sign.
Coulomb's Law tells you that
19.3 N = k*Q1*Q2/R^2, where R = 1.1 m
Substitute 505*10^-6 - Q1 for Q2 and solve for Q1.
You will have to look up k. It is something like 8.99*10^9 N(m^2/C^2), but I don't trust my memory.
Use Coulomb's law:
F=(1/(4πε0))*(q1*q2/r²)
where
ε0=8.85*10^-12 C² N-1 m-2
q1,q2 are charges (C)
r=distance between the charges (m)
The equation applies when the distance r is large compared to the size of the charged particles.
The force is repulsive when positive, and negative when attractive.
Thus if one charge is q, the other charge would be .000505-q.
We have
F=q(.000505-q)/(4π&epsilon0)
where all quantities are known except q.
Solve for q.
F=(1/(4πε0))*(q1*q2/r²)
where
ε0=8.85*10^-12 C² N-1 m-2
q1,q2 are charges (C)
r=distance between the charges (m)
The equation applies when the distance r is large compared to the size of the charged particles.
The force is repulsive when positive, and negative when attractive.
Thus if one charge is q, the other charge would be .000505-q.
We have
F=q(.000505-q)/(4π&epsilon0)
where all quantities are known except q.
Solve for q.
1 answer