The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2) where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.
Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span>
The elapsed time when the particle returns to the origin is determined from the ratio of initial velocity and acceleration of the particle.
<h3>Time of motion of the particle</h3>
The time of motion of the particle is calculated by applying Newton's second law of motion.
F = ma
F = m(v)/t
where;
- t is time of motion of the particle
- m is mass of the particle
- v is velocity of the particle
a = v - u/t
v = u + at
when the particle returns to the origin, direction of u, = negative.
final velocity = 0
0 = -u + at
at = u
t = u/a
Learn more about force here: brainly.com/question/12970081
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Answer:
Electric potential, E = 2100 volts
Explanation:
Given that,
Electric field, E = 3000 N/C
We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m
The electric potential is given by :
V = 2100 volts
So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.
Answer:
abiotic
Explanation:
i think but dont take my word for it
