Answer:
V = 576 V
Explanation:
Given:
- The area of the two plates A = 0.070 m^2
- The space between the two plates d = 6.3 mm
- Te energy density u = 0.037 J /m^3
Find:
- What must the potential difference between the plates V?
Solution:
- The energy density of the capacitor with capacitance C and potential difference V is given as:
u = 0.5*ε*E^2
- Where the Electric field strength E between capacitor plates is given by:
E = V / d
Hence,
u = 0.5*ε*(V/d)^2
Where, ε = 8.854 * 10^-12
V^2 = 2*u*d^2 / ε
V = d*sqrt ( 2*u / ε )
Plug in values:
V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )
V = 576 V
Answer:
Gas
Explanation:
Cos Milky Said it. Milky Smart.
Answer:
Pish posh idk xD im busy but ill help in a second
Explanation:
Answer:
Explanation:
give data:
inside diameter = 5.0 cm
charge q = 0.25 nC
Outside diameter = 15 cm
potential V at inside sphere is = 
potential V at outside sphere is = 
k is constant whose value is = 
then potential difference between two point is
![\Delta V = kq \left [\frac{1}{R}-\frac{1}{r} \right ]](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20kq%20%5Cleft%20%5B%5Cfrac%7B1%7D%7BR%7D-%5Cfrac%7B1%7D%7Br%7D%20%20%5Cright%20%5D)
![\Delta V = 9*10^{9}*0.25*10^{-9} \left [\frac{1}{0.05}-\frac{1}{0.15} \right ]](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%209%2A10%5E%7B9%7D%2A0.25%2A10%5E%7B-9%7D%20%5Cleft%20%5B%5Cfrac%7B1%7D%7B0.05%7D-%5Cfrac%7B1%7D%7B0.15%7D%20%5Cright%20%5D)
