The electric field generated by a point charge is given by:
![E= k_e \frac{Q}{r^2}](https://tex.z-dn.net/?f=E%3D%20k_e%20%5Cfrac%7BQ%7D%7Br%5E2%7D%20)
where
![k_e = 8.99 \cdot 10^9 Nm^2 C^{-2}](https://tex.z-dn.net/?f=k_e%20%3D%208.99%20%5Ccdot%2010%5E9%20Nm%5E2%20C%5E%7B-2%7D)
is the Coulomb's constant
Q is the charge
r is the distance from the charge
We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them.
Let's calculate first the electric field generated by the positive charge at that point:
![E_1=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C](https://tex.z-dn.net/?f=E_1%3Dk_e%20%20%5Cfrac%7BQ_1%7D%7Br%5E2%7D%3D%288.99%20%5Ccdot%2010%5E9%20Nm%5E2C%5E%7B-2%7D%29%20%5Cfrac%7B%282.0%20%5Ccdot%2010%5E%7B-6%7D%20C%29%7D%7B%280.05%20m%29%5E2%7D%20%3D%2B7.19%20%5Ccdot%2010%5E6%20N%2FC%20)
where the positive sign means its direction is away from the charge.
while the electric field generated by the negative charge is:
![E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C](https://tex.z-dn.net/?f=E_2%3Dk_e%20%5Cfrac%7BQ_1%7D%7Br%5E2%7D%3D%288.99%20%5Ccdot%2010%5E9%20Nm%5E2C%5E%7B-2%7D%29%20%5Cfrac%7B%28-2.0%20%5Ccdot%2010%5E%7B-6%7D%20C%29%7D%7B%280.05%20m%29%5E2%7D%20%3D-7.19%20%5Ccdot%2010%5E6%20N%2FC%20)
where the negative sign means its direction is toward the charge.
If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
Answer:
hello your question is incomplete attached below is the complete question
answer : The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor i.e. J > Jm
Explanation:
Gear ratio G > 1
a) Determine the moment of inertia felt by the motor
moment of inertia felt by Motor = moment of Inertia at the armature
b) Determine the moment of inertial felt by someone who is rotating the mass by hand
moment of inertia felt by someone is = J
The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor
attached below is a detailed solution
Answer:
The mass of the object involved and the value of the gravitational acceleration
Explanation:
- Gravitational potential energy is defined as the energy possessed by an object in a gravitational field due to its position with respect to the ground:
![U=mgh](https://tex.z-dn.net/?f=U%3Dmgh)
where m is the mass of the object, g is the gravitational acceleration and h is the heigth of the object with respect to the ground.
- Elastic potential energy is defined as the energy possessed by an elastic object and it is given as:
![U=\frac{1}{2}kx^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
where k is the spring constant of the elastic object, while x is the compression/stretching of the spring with respect to the equilibrium position.
As we can see from the equations, both types of energy depends on the relative position of the object/end of the spring with respect to a certain reference position (h in the first formula, x in the second formula), but gravitational potential energy also depends on m (the mass) and g (the gravitational acceleration) while the elastic energy does not.
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