Answer:
Electric force on the charge will be equal to -1742000 N
Explanation:
We have given electric field at a point is E = 260000 N/C
And charge q = -6.7 C
We have to find the force on that charge
Force on the charge is equal to Force = charge × electric field ( multiplication of electric field and charge )
So force will be equal to 
So electric force on the charge will be equal to -1742000 N
Answer:
1.) Time t = 3.1 seconds
2.) Height h = 46 metres
Explanation:
given that the initial velocity U = 30 m/s
At the top of the trajectory, the final velocity V = 0
Using first equation of motion
V = U - gt
g is negative 9.81m/^2 as the object is going against the gravity.
Substitute all the parameters into the formula
0 = 30 - 9.81t
9.81t = 30
Make t the subject of formula
t = 30/9.81
t = 3.058 seconds
t = 3.1 seconds approximately
Therefore, it will take 3.1 seconds to reach to reach the top of its trajectory.
2.) The height it will go can be calculated by using second equation of motion
h = ut - 1/2gt^2
Substitutes U, g and t into the formula
h = 30(3.1) - 1/2 × 9.8 × 3.1^2
h = 93 - 47.089
h = 45.911 m
It will go 46 metres approximately high.
Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance.
