Answer:
1408.685 KN/C
Explanation:
Given:
R = 0.45 m
σ = 175 μC/m²
P is located a distance a = 0.75 m
k = 8.99*10^9
- The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

part a)
Electric Field strength at point P: a = 0.75 m

part b)
Since, R >> a, we can approximate a / R = 0 ,
Hence, E simplified relation becomes:

E = σ / 2*e_o
part c)
Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:
Electric Field strength due to point charge is:
E = k*δ*pi*R^2 / a^2
Since, R << a, Surface area = δ*pi
Hence,
E = (k*δ*pi/a^2)
Answer:
a=2500J,b=1000K,c=1000J,d=14.142m/s
Explanation:
V²=U²+2gh
V²=0 + 2×10×10=200m/s
a).kinetic energy=(1/2)mv²=(1/2)25×200=2500
potential energy=mgh
p.e=25×10×10=2500J
pe+ke=2500+2500=5KJ
b).mgh=25×10×4=1000J
c). V²=U²+2gh
V²=0+2×10×4
V²=80
kinetic energy=(1/2)mv²
=(1/2)25×80
=1KJ
d). From my first paragraph V²=200
V=√200
V=14.142m/s