Explanation:
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
Answer:
3.536*10^-6 C
Explanation:
The magnitude of the charge is expresses as Q = CV
C is the capacitance of the capacitor
V is the voltage across the capacitor
Get the capacitance
C = ε0A/d
ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m
A is the area = 0.2m²
d is the plate separation = 0.1mm = 0.0001m
Substitute
C = 8.84 x 10-12 * 0.2/0.0001
C = 1.768 x 10-8 F
Get the potential difference V
Using the formula for Electric field intensity
E = V/d
2.0 × 10^6 = V/0.0001
V = 2.0 × 10^6 * 0.0001
V = 2.0 × 10^2V
Get the charge on each plate.
Q = CV
Q = 1.768 x 10-8 * 2.0 × 10^2
Q = 3.536*10^-6 C
Hence the magnitude of the charge on each plate should be 3.536*10^-6 C
Explanation:
If a positive test charge is placed in an electric field, it will exert the force in the test charge in the direction of electric field vector. We know that the direction of electric field is given by electric field lines. The field lines for a positive charge is outwards. The electric force acting on the charge is given by :
F = q E
Hence, this is the required solution.
Answer: It frees up valuable portions of the broadcast spectrum, it has better audio and picture quality, and there are more options on digital broadcasting
Explanation:
I believe the acceleration would be 5m/s
All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.
