The Electric field is zero at a distance 2.492 cm from the origin.
Let A be point where the charge
C is placed which is the origin.
Let B be the point where the charge
C is placed. Given that B is at a distance 1 cm from the origin.
Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.
i.e., at distance 'x' from B.
Using Coulomb's law,
,
= 



k is the Coulomb's law constant.
On substituting the values into the above equation, we get,

Taking square roots on both sides and simplifying and solving for x, we get,
1.67x = 1+x
Therefore, x = 1.492 cm
Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.
Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926
#SPJ4
As we know that when charge is released in electric field
It will have two forces on it
1. electrostatic force
2. gravitational force
now if the ball will accelerate upwards so we can say
net upward force = mass * acceleration


now we can find charge q on it by above equation

So above is the charge on the particle
m = mass of the car moving in horizontal circle = 1750 kg
v = Constant speed of the car moving in the horizontal circle = 15 m/s
r = radius of the horizontal circular track traced by the car = 45.0 m
F = magnitude of the centripetal force acting on the car
To move in a circle . centripetal force is required which is given as
F = m v²/r
inserting the above values in the formula
F = (1750) (15)²/(45)
F = (1750) (225)/(45)
F = 1750 x 5
F = 8750 N