The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.
Location ' x ' is √(2² + 3²) = √13 m from the charge.
Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.
The magnitude of the E-field is the same at both locations.
The direction is also the same at both locations ... it points toward the origin.
Answer:
Work done, W = 0.0219 J
Explanation:
Given that,
Force constant of the spring, k = 290 N/m
Compression in the spring, x = 12.3 mm = 0.0123 m
We need to find the work done to compress a spring. The work done in this way is given by :


W = 0.0219 J
So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.
Answer:
E_total = 3 N / A
Explanation:
The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.
E_total = E + E₁
the expression for the field of a point charge is
E₁ = k q₁ / r²
for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative
E_total = E -k q₁ / r₂
we substitute
0 = E - k q₁ / r²
q₁ =
let's calculate
q₁ =
q₁ = 1.78 10⁻⁹ C
now we can calculate the field for position x = 4 m
E_total = 4 - 9 10⁹ 1.78 10⁻⁹ / 4²2
E_total = 3 N / A
Answer:
A. 12,240.
Explanation:
1,530 times 8.0 = 12,240.
hope this helped :)
Answer:
r = 1.45 Å
Explanation:
given,
λ = 1.436 Å
θ = 20.62°
d = a
n = 2
metal gold crystallizes in a face centered cubic unit cell
Radius of the gold atom = ?
using Bragg's Law
n λ = 2 d sin θ
2 x 1.436 Å = 2 a sin 20.62°
a = 4.077 Å
We know relation of radius for face centered cubic unit cell


r = 1.45 Å
the radius of a(n) gold atom. is equal to 1.45 Å