The answer is A. the fields lines never cross, if you bring another magnet near it, the lines work just compress
Answer:
![x=\dfrac{r}{\sqrt2}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7Br%7D%7B%5Csqrt2%7D)
Explanation:
Given that
Radius =r
Electric filed =E
Q=Charge on the ring
The electric filed at distance x given as
![E=K\dfrac{Q}{(r^2+x^2)^{3/2}}](https://tex.z-dn.net/?f=E%3DK%5Cdfrac%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B3%2F2%7D%7D)
For maximum condition
![\dfrac{dE}{dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7BdE%7D%7Bdx%7D%3D0)
![E=K{Q}{(r^2+x^2)^{-3/2}}](https://tex.z-dn.net/?f=E%3DK%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-3%2F2%7D%7D)
![\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}](https://tex.z-dn.net/?f=%5Cdfrac%7BdE%7D%7Bdx%7D%3DK%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-3%2F2%7D%7D-%5Cdfrac%7B3%7D%7B2%7D%5Ctimes%202%5Ctimes%20x%5Ctimes%20K%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-5%2F2%7D%7D)
For maximum condition
![\dfrac{dE}{dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7BdE%7D%7Bdx%7D%3D0)
![K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0](https://tex.z-dn.net/?f=K%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-3%2F2%7D%7D-%5Cdfrac%7B3%7D%7B2%7D%5Ctimes%202%5Ctimes%20x%5Ctimes%20K%7BQ%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B-5%2F2%7D%7D%3D0)
![r^2+x^2-3x^2=0](https://tex.z-dn.net/?f=r%5E2%2Bx%5E2-3x%5E2%3D0)
![x=\dfrac{r}{\sqrt2}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7Br%7D%7B%5Csqrt2%7D)
At
the electric field will be maximum.
Answer:
the force will decrease to 3/4 of its original value.
Explanation:
The initial electric force between the two charges is:
![F = k \frac{q\cdot q}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20k%20%5Cfrac%7Bq%5Ccdot%20q%7D%7Br%5E2%7D)
where
k is the Coulomb's constant
q is the magnitude of each charge
r is their separation
Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of
![q+\frac{q}{2}=\frac{3}{2}q](https://tex.z-dn.net/?f=q%2B%5Cfrac%7Bq%7D%7B2%7D%3D%5Cfrac%7B3%7D%7B2%7Dq)
while the other charge will be
![q-\frac{q}{2}=\frac{q}{2}](https://tex.z-dn.net/?f=q-%5Cfrac%7Bq%7D%7B2%7D%3D%5Cfrac%7Bq%7D%7B2%7D)
So, the new force will be
![F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F](https://tex.z-dn.net/?f=F%27%20%3D%20k%20%5Cfrac%7B%28%5Cfrac%7Bq%7D%7B2%7D%29%5Ccdot%20%28%5Cfrac%7B3%7D%7B2%7Dq%29%7D%7Br%5E2%7D%3D%5Cfrac%7B3%7D%7B4%7D%20%28k%5Cfrac%7Bq%5Ccdot%20q%7D%7Br%5E2%7D%29%3D%5Cfrac%7B3%7D%7B4%7DF)
So, the force will decrease to 3/4 of its original value.
Given :
Initial speed of car A is 15 m/s and initial speed of car B is zero.
Final speed of car A is zero and final speed of car B is 10 m/s.
To Find :
What fraction of the initial kinetic energy is lost in the collision.
Solution :
Initial kinetic energy is :
![K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}](https://tex.z-dn.net/?f=K.E_i%20%3D%20%5Cdfrac%7B15%5E2m%7D%7B2%7D%20%2B%200%5C%5C%5C%5CK.E_i%20%3D%20%5Cdfrac%7B225%20m%7D%7B2%7D)
Final kinetic energy is :
![K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}](https://tex.z-dn.net/?f=K.E_f%20%3D%20%5Cdfrac%7B10%5E2m%7D%7B2%7D%20%2B%200%5C%5C%5C%5CK.E_f%20%3D%20%5Cdfrac%7B100m%7D%7B2%7D)
Now, fraction of initial kinetic energy loss is :
![Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25](https://tex.z-dn.net/?f=Loss%20%3D%20%5Cdfrac%7B%5Cdfrac%7B225m%7D%7B2%7D-%5Cdfrac%7B100m%7D%7B2%7D%7D%7B%5Cdfrac%7B100m%7D%7B2%7D%7D%5C%5C%5C%5CLoss%20%3D%20%5Cdfrac%7B125%7D%7B100%7D%5C%5C%5C%5CLoss%20%3D%201.25)
Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .
green liquid becoming a red liquid