Answer:
Sorry I'm wrong The person above is correct.
I tried.
Answer:
x=0.53
Explanation:
Using Gauss law the field is uniform so
E=ζ/ε
Charge densities ⇒ζ=1.
ε=8.85

Force of charge is


So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

Answer:

Explanation:
We can assume this problem as two concentric spherical metals with opposite charges.
We have also to take into account the formulas for the electric field and the capacitance. Hence we have

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating
![dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]](https://tex.z-dn.net/?f=dV%3DEdr%5C%5C%5C%5CV%3D%5Cint_%7BR_1%7D%5E%7BR_2%7DEdr%3D-%5Cint_%7BR_1%7D%5E%7BR_2%7D%5Cfrac%7BkQ%7D%7Br%5E2%7Ddr%5C%5C%5C%5CV%3DkQ%5B%5Cfrac%7B1%7D%7BR_2%7D-%5Cfrac%7B1%7D%7BR_1%7D%5D)
Hence, the capacitance is
![C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B1%7D%7Bk%5B%5Cfrac%7B1%7D%7BR_2%7D-%5Cfrac%7B1%7D%7BR_1%7D%5D%7D)
but R1=a and R2=b

HOPE THIS HELPS!!
Answer:
act or process of measuring
Explanation: