The answer is c. the degree of wetness of the paper towels.
Answer:
V = 576 V
Explanation:
Given:
- The area of the two plates A = 0.070 m^2
- The space between the two plates d = 6.3 mm
- Te energy density u = 0.037 J /m^3
Find:
- What must the potential difference between the plates V?
Solution:
- The energy density of the capacitor with capacitance C and potential difference V is given as:
u = 0.5*ε*E^2
- Where the Electric field strength E between capacitor plates is given by:
E = V / d
Hence,
u = 0.5*ε*(V/d)^2
Where, ε = 8.854 * 10^-12
V^2 = 2*u*d^2 / ε
V = d*sqrt ( 2*u / ε )
Plug in values:
V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )
V = 576 V
Answer:
F = 84.61 N
Explanation:
As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.
component of Force along the incline = F cos(23° - 15°) = F cos(8°)
component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)
Equating the above two components of forces will give the minimum Force required.
F cos(8°) = 33*9.81*sin(15°)
F = 33*9.81*sin(15°) / cos(8°) (calculate the value using a scientific calculator)
<u>F = 84.61 N</u>
The correct answer is:
sixteen times
In fact, the distance between charge q and the source S is 1 unit. Instead, the distance between charge p and the source S is 4 units. The magnitude of the electrostatic force is inversely proportional to the square of the distance between the charge and the source:

where r is the distance. If we take the force between q and S as reference, we have r=1, so that

while the force between p and S is

Therefore, we see that the force exerted between q and S is 16 times the force exerted between p and S.