Answer:
0.018 J
Explanation:
The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

where
is the magnitude of the charge
is the potential difference between point P and infinity
Substituting into the equation, we find

Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are


where
is the initial velocity.
(a).
When the projectile hits the 50m mark,
; therefore,

solving for
we get:

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

which gives

(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

the vertical component of the velocity is

which gives a speed
of


We need the frequency of the photon, it is v = c/ λ
Where c is 3 x 10^8 ms^-1 and λ
is the wave length
We also need the expression of
connecting frequency to energy of photon
which is E = hv where h is Planck’s
constant
Combining the two equations
will give us:
E = h x c/λ
Inserting the values, we will
have:
E = 6.626 x 10^-34 x 3 x 10^8 /
0.126
E = 1.578 x 10^ -24 J
Answer:
A). σ = 3.823 x
/N-
B).
C/
C).
J
Explanation:
A). We know magnitude of charge per unit area for a conducting plate is given by

where, E is resultant electric field = 1.2 x
V/m
is permittivity of free space = 8.85 x
/N-
k is dielectric constant = 3.6
∴
= 3.6 x 8.85 x
x 1.2 x 
= 3.823 x
/N-
B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by


C/
C).
Area of the plate, A = 2.5 
= 2.5 x 

diameter of the plate, d = 1.8 mm
= 1800 m
∴ Total energy stored in the capacitor


J
The angle of refraction would be further less