A string Im long is fixed at one end. The
other end is moved up and down with
frequency 15 Hz. Due to this, a stationary
wave with four complete loops, gets
produced on the string. Find the speed of
the progressive wave which produces the
stationary wave (Hint: Remember that
the moving end is an antinode.]
Answers
Explanation:
The inverse-square nature of both laws is just a consequence of the three-dimensional nature of space:
The surface through which the field flows at the same distance r from the source, is a sphere, and the surface area of that sphere is given by A = 4π r². [Which in turn can be derived by integrating over the surface in spherical coordinates.]
Consider Gauss’ law applied to the electric field
∫_A E⃗ · dA⃗ = 1/ε₀ ∑_A q_i
∫_A |E⃗| |dA⃗| cos(∠(E⃗, dA⃗)) = 1/ε₀ ∑_A q_i,
(“the electric field through a surface is proportional to the electric charge contained by it”), where E⃗ is the electric field through the surface with area A given by the normal vector dA⃗, ε₀ is the electric (field proportionality) constant, and the q_i are individual electric charges, so ∑_A q_i is the total electric charge contained by the surface.
The sum of all contained charges is the total charge Q:
∫_A |E⃗| |dA⃗| cos(∠(E⃗, dA⃗)) = 1/ε₀ Q.
The electric field lines are perpendicular to the surface, so the angle between the field vector and the surface normal vector is zero (0 ≡ 0°). The cosine of 0 is 1:
∫_A |E⃗| |dA⃗| cos(0) = 1/ε₀ Q
∫_A |E⃗| |dA⃗| = 1/ε₀ Q.
The electric field strength is the same everywhere at the surface, so we can eliminate it from the integral:
|E⃗| ∫_A |dA⃗| = 1/ε₀ Q
E ∫_A dA = 1/ε₀ Q.
dA = |dA⃗| = |du⃗ × dv⃗| is the area of an infinitesimal surface element given by the spatial dimensions u and v along the surface [the magnitude of the cross product of two vectors is equal to the area of the parallelogram produced by them: |a⃗ × b⃗| = |a⃗| |b⃗| sin(∠(a⃗, b⃗))], so if we integrate that over the surface, we just get the total surface area A:
E A = 1/ε₀ Q.
The surface area at equal distance r from the source is that of a sphere (AISB):
E (4π r²) = 1/ε₀ Q.
So the electric field strength at distance r is
E(r) = 1/ε₀ Q/(4π r²) = 1/(4π ε₀) Q/r².
By Gauss’ law (which also applies to gravitation as it was initially formulated for that) it does not matter for a given volume how the electric charge (with gravity: the mass) is distributed within that volume: We can assume that all of the electric charge in that volume is concentrated in one point (point charge; with gravity: point mass).
So the electric field strength at distance r from a point charge q₁ is also
E(r) = 1/(4π ε₀) q₁/r².
The strength of the electric field of q₁ was defined as a measure of force on an infinitesimal test charge q₂:
E(r) = lim_{q₂ → 0} F(r)/q₂ = 1/(4π ε₀) q₁/r²
So the force between two point charges is
F(r) = 1/(4π ε₀) q₁q₂/r².