Explanation:
It is given that,
Magnitude of charge, 
It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.
Magnetic field, 
Velocity, 
![v=[(4.53)i+(2.11)j]\ m/s](https://tex.z-dn.net/?f=v%3D%5B%284.53%29i%2B%282.11%29j%5D%5C%20m%2Fs)
We need to find the magnitude of force on the charge. Magnetic force is given by :

![F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]](https://tex.z-dn.net/?f=F%3D15%5Ctimes%2010%5E%7B-6%7D%5B%284.53i%2B2.11j%29%5Ctimes%200.08%5C%20j%5D)
<em>Since</em>, 
![F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]](https://tex.z-dn.net/?f=F%3D15%5Ctimes%2010%5E%7B-6%7D%5B%284.53i%29%5Ctimes%20%280.08%29%5C%20j%5D)


So, the force acting on the charge is
and is moving in positive z axis. Hence, this is the required solution.
Answer:
The capacitance is 11 F for half and fully charged capacitor.
Explanation:
Capacitance of capacitor is given by the expression

Where ε is the vacuum permittivity, A is the area of plates and d is the separation between plates.
So capacitance does not depend upon charge and potential. So capacitance fully and half charged capacitors are same.
Here the capacitance is 11 F for half and fully charged capacitor.
Answer:
12.0 V
Explanation:
Data :
Potential difference due to a single charge (+Q), E = 3.0 V
The Electric potential for the system of charges is given as:
![E=\frac{1}{4\pi \epsilon_o}[\Sigma\frac{Q}{r}]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_o%7D%5B%5CSigma%5Cfrac%7BQ%7D%7Br%7D%5D)
for single charge, E = 3.0 V =
->eq(1)
And for 4 charges:
-eq(2)
from eq(1) and (2) we have
E = 4 × 3.0 V = 12 V
Answer:
Friction does not oppose motion, it oppose relative motion.
Explanation:
static friction opposes relative motion
<span>they had originally used einsteins therory of relativity and his field equations to figure out their information</span>