if a particle with a charge of +4.3 C is attracted to another particle by a force of 6.5 N, what is the strength of the electric field at this location?

1.5 N/C

Explanation:

The equation is derived from Coulomb's Law, but is not exactly Coulomb's Law:

F = qE

q is the test charge

E is the electrical field produced by the source charge

F is the force in Newtons

In this problem, we were given the value of the test charge q, and the force acting on it due to the electric field from the source charge.

E = F/q

E = 6.5N/4.3C

E = 1.5 N/C

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft