The expression of the electric flux is
Here,
Q = Total charge enclosed in the closed surface
= Permittivity due to free space
Rearranging to find the charge,
Replacing with our values we have finally
The charge enclosed by the box is 0.1684nC
The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.
The first choice on the list is the correct one.
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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Answer:
Exercise 1;
The centripetal acceleration is approximately 94.52 m/s²
Explanation:
1) The given parameters are;
The diameter of the circle = 8 cm = 0.08 m
The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m
The speed of motion = 7 km/h = 1.944444 m/s
The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²
The centripetal acceleration ≈ 94.52 m/s²
Answer: The Electrostatic force of attraction or repulsion between two charges shows that the Newton's third law applies to electrostatic forces.
Explanation: Consider two Oppositely charged charges separated by distance d.
The electrostatic force exerted by charge 1 on charge 2 is.
By Coulomb's Law :
F1 = k 
.....................................(1)
The electrostatic force exerted by charge 2 on charge 1 is.
F2 = - k ................................. (2)
negative sign shows that force are in opposite direction.
From Equation 1 and 2
F1 = - F2
Which implies Newton Third law.
