V= 274 meters / 23 sec
V= 11.91 meters per sec
Answer:
The ratio (U₁/U₂) = 6
Explanation:
U, the potential energy is given as
U = kqQ/r
k = Coulomb's constant
q = charge we're concerned about
Q = charge of the negative plate of the capacitor
r = distance of q from the negative plate of the capacitor.
For charge q₁
U₁ = kq₁Q/s
U₂ = kq₂Q/2s
But q₂ = q₁/3
U₂ becomes U₂ = kq₁Q/6s
U₁ = kq₁Q/s
U₂ = kq₁Q/6s
(U₁/U₂) = 6
Answer:
q = - 93.334 nC
Explanation:
GIVEN DATA:
Radius of ring 73 cm
charge on ring 610 nC
ELECTRIC FIELD p FROM CENTRE IS AT 70 CM
E = 2000 N/C
Electric field due tor ring is guiven as
![E = \frac{KQx}{[x^2+ R^2]^{3/2}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BKQx%7D%7B%5Bx%5E2%2B%20R%5E2%5D%5E%7B3%2F2%7D%7D)

E1 = 3714.672 N/C
electric field due to point charge q



now the eelctric charge at point P is
E = E1 + E2
solving for q
q = - 93.334 nC
Answer:
V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)
Explanation:
Attached is the full solution