Answer:
its potential energy decreases and its electric potential decreases.
Explanation:
Let's consider a radial field for simplicity. We have:
- The electric potential of the field is given by:
where
k is the Coulomb's constant
Q is the charge source of the field
r is the distance from Q
We see that the electric potential decreases as we move away from the source. If we consider a positive charge q moving in the direction of the electric field, this charge q will move away from the charge Q (because the field lines generated by the positive particle Q point away from the particle), so the electric potential will decrease.
- The potential energy of the moving charge is given by
where q is the magnitude of the charge. As we said previously, V is decreasing while the charge is moving in the direction of the field, so since U is directly proportional to V, U will decrease as well.
F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁