Answer:
![V_1=8 V_2](https://tex.z-dn.net/?f=V_1%3D8%20V_2)
Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2,
![d_2=d](https://tex.z-dn.net/?f=d_2%3Dd)
- separation distance of capacitor 1,
![d_1=2d](https://tex.z-dn.net/?f=d_1%3D2d)
- quantity of charge on capacitor 2,
![Q_2=Q](https://tex.z-dn.net/?f=Q_2%3DQ)
- quantity of charge on capacitor 1,
![Q_1=4Q](https://tex.z-dn.net/?f=Q_1%3D4Q)
We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
.....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air
![\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}](https://tex.z-dn.net/?f=%5Cepsilon_0%20%3D%208.85%5Ctimes%2010%5E%7B-12%7D%5C%2CF.m%5E%7B-1%7D)
From eq. (1)
For capacitor 2:
![C_2=\frac{k.\epsilon_0.A}{d}](https://tex.z-dn.net/?f=C_2%3D%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D)
For capacitor 1:
![C_1=\frac{k.\epsilon_0.A}{2d}](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7B2d%7D)
![C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B1%7D%7B2%7D%20%5B%20%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%5D)
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:
![V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BQ%7D%7B%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%7D)
![V_2=\frac{Q.d}{k.\epsilon_0.A}](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D)
& for capacitor 1:
![V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}](https://tex.z-dn.net/?f=V_1%3D%5Cfrac%7B4Q%7D%7B%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7B2d%7D%7D)
![V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}](https://tex.z-dn.net/?f=V_1%3D%5Cfrac%7B4Q%5Ctimes%202d%7D%7Bk.%5Cepsilon_0.A%7D)
![V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]](https://tex.z-dn.net/?f=V_1%3D8%5Ctimes%20%5B%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D%5D)
![V_1=8 V_2](https://tex.z-dn.net/?f=V_1%3D8%20V_2)
Answer:
B. 175 N
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
Where;
Fnet is the net force
Fapp is the applied force
Fg is the force due to gravitation
In this scenario, we observed that both forces are acting in the same direction.
Therefore:
Net force = 100 N + 75 N
Net force = 175 Newton
Answer:
A. Remove everything in the refrigerator to lighten the load.
B. Put a lubricant between the surface of the object and the floor
C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily
Explanation:
Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.
F = μN where N is the normal force which depends on the mass.
Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.
D
Because the rest of the answers are illogical