What is the universe’s estimated age?
2 answers:
You might be interested in
Answer:
Figure a. E_net = 99.518 N/C
Figure b. E_net = 177.151 N / C
Explanation:
Given:
- Attachment for figures missing in the question.
- The dimensions for rectangle are = 7.79 x 3.99 cm
- All four charges have equal magnitude Q = 10.6*10^-12 C
Find:
Find the magnitude of the electric field at the center of the rectangle in Figures a and b.
Solution:
- The Electric field generated by an charged particle Q at a distance r is given by:
E = k*Q / r^2
- Where, k is the coulomb's constant = 8.99 * 10^9
Part a)
- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:
E_1 + E_3 = 0
- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:
E_net = E_2 + E_4
E_2 = E_4
E_net = 2*E = 2*k*Q / r^2
- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:
r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )
r = sqrt ( 1.9151*10^-3 ) = 0.043762 m
- Plug the values in the E_net expression developed above:
E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3
E_net = 99.518 N/C
Part b)
- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).
- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.
Hence,
E_net = 2*E_part(a)*cos(Q)
- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:
Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.
- Now, compute the net electric field E_net:
E_net = 2*(99.518)*cos(27.12)
E_net = 177.151 N / C
Answer:
The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N
Explanation:
The question relates to the force of gravity experienced between two bodies
The given parameters are;
The mass of Big Ben, M₁ = 1 × 10⁸ kg
The mass of the Empire State Building, M₂ = 1 × 10⁹ kg
The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters
By Newton's Law of gravitation, we have;
Where;
F = The force exerted by Big Ben on the Empire State Building and vice versa
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
M₁, M₂, and r are the given parameters
By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;
The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.