Comment
The only reason you can do this is that the charges are the same. If they were not, the problem would not be possible.
Equation
The field equation is, in its simplest form,
E = kq/r^2
So each of the charges are pulling / pushing in the same direction. The equation becomes.
kq/r^2 - (-kq/r^2) = Field magnitude in N/C
Givens
- K = 9 * 10^9 N m^2 / c^2
- E = 45 N/C
- r = 7.5/2 = 3.75 cm * ( 1 m / 100 cm) = 0.0375 m
- Find Q
Solution
k*q/0.0375 ^2 - (-kq/0.0375^2) = 45 N/C Combine
2*k*q / 0.0375^2 = 45 N/C Divide by 2
kq /(0.0375^2) = 22.5 N/C Multiply by 0.0375^2
kq = 22.5 * 0.0375 ^2 Find d^2
kq = 22.5 * 0.001406 Combine
kq = 0.03164 N/C * m^2 Divide by k
q = 0.03164 N * m^2 /C / 9*10^9 N m^2 / c^2
q = 2.84760 * 10 ^8 C
I've left the cancellation of the units for you. Notice that only 1 C is left and it is in the numerator as it should be.
Answer:
The answer is C. Its not on mars and its not 10 football feilds its only 1
Answer:
vi = 2.83 √gR
Explanation:
For this exercise we can use the law of conservation of energy
Let's take a reference system that is at point A, the lowest
Starting point. Lower, point A
Em₀ = Ki = ½ m vi²
Final point. Higher, point B
= K + U
It indicates that at this point the kinetic energy is ki / 2 and the potential energy is ki / 2
K = ki / 2
U = m g (2R)
Energy is conserved so
Em₀ = Em_{f}
½ m vi² = ½ (1/2 m vi²) + m g 2R
½ m vi² (1- ½) = m g 2R
vi² = 4 g 2 R
vi = √ 8gR = 2 √2gR
vi = 2.83 √gR
